Answer: θ=0.75448...πn
Use the Law of Cosines to determine the indicated angle theta.
(Assume a = 122.5, b = 58.3, and c = 164.5. Round your answer to
the nearest degree.)
The angle opposite to the side length c is given as follows:
θ = 53º.
What is the law of cosines?The Law of Cosines is a trigonometric formula that relates the lengths of the sides of a triangle to the cosine of one of its angles. It is also known as the Cosine Rule.
The Law of Cosines states that for any triangle with sides a, b, and c and angle C opposite to side c, the following equation holds true:
c^2 = a^2 + b^2 - 2ab cos(C)
The parameters for this problem are given as follows:
a = 122.5, b = 58.2, c = 164.5.
Hence the measure of angle θ is given as follows:
c^2 = a^2 + b^2 - 2ab cos(θ)
164.5² = 122.5² + 58.2² - 2 x 122.5 x 58.2 x cos(θ)
14259cos(θ) = 8666.76
cos(θ) = 8666.76/14259
θ = arccos(8666.76/14259)
θ = 53º.
Missing InformationThe angle is opposite to the side length C.
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Scientists captured, tagged, and released 20 crows as part of a research study. A week later, they counted 250 crows, of which 10 had tags. To the nearest whole number, what is the best estimate for the crow population?
Rewrite as expression. Twenty divided by a number
can someone show me how to do this
Answer: (0, 2)
Step-by-step explanation:
The solution of a system of equations is where the lines intersect at.
We will use substitution to solve the system.
Equations:
y = 1/2x + 2
y = -1/5x + 2
Set both equations to equal each other:
1/2x + 2 = -1/5x + 2
Simplify:
7/10x = 0
x = 0
Plug 0 back in:
y = 1/2(0) + 2
y = 0 + 2
y = 2
The solution is (0, 2)
(This can also be seen by looking at the graph)
Hope this helps!
I need help ASAP! What is the expanded form of this number
15.306
A. (1 x 10) + (5 x 1) + (3 x 1/10) + (6 x 1/1000)
B. (1×10) + (5×1) + (3×110) + (6× 1/1000)
C. (1×10) + (5×1) + (3×1/100) +(6×1/100)
D. (1×10) + (5×1) + (3×1/100) + (6×1/1,000)
Answer:
the answer is A
So basically it's simple just use some scientific calculator and put the number one by one with the bracket because there's a rule of mathematics which we call BODMAS
20 POINTSS!! HURRY!
Airplane A traveled 1662 miles in six hours.
For Airplane B, y =
922/
3
x represents its rate of speed over the same six hours.
Compare the two airplanes to determine which airplane traveled at a faster rate?
Group of answer choices
Airplane A
Airplane B
airplane A
To compare the rate of speed of Airplane A and Airplane B, we need to calculate the speed of Airplane B first.
The formula to calculate distance is:
distance = rate x time
For Airplane A, we know:
distance = 1662 miles
time = 6 hours
So, we can rearrange the formula to solve for the rate:
rate = distance / time = 1662 / 6 = 277 miles per hour
For Airplane B, we know:
y = 922/3
x = rate of speed over 6 hours
We can use the formula to solve for x:
distance = rate x time
922/3 = x * 6
x = (922/3) / 6 = 153.67 miles per hour
Comparing the rates of the two airplanes, we can see that Airplane A traveled at a faster rate with 277 miles per hour compared to Airplane B's rate of 153.67 miles per hour. Therefore, Airplane A traveled at a faster rate.
How long would Debbie have to leave her money in two accounts, $5000 at 8.75% and $7000 at
9.5%, in order to earn $8820 in interest?
Assuming this is a simple interest situation:
The 8.75% account will earn (0.0875)(5000) each year.
$437.5 earned per year
The 9.5 % account will earn (0.095)(7000) each year
$665 earned per year
Together, these accounts earn $1102.5 per year.
If x is the number of years until the desired interest is earned, then you need to solve 1102.5x = 8820. Dividing by 1102.5 on both sides, you find x=8.
Again, assuming this is simple interest and not compound interest, it will take 8 years.
A population of bacteria is decaying.
The number of bacteria after h hours is given by the expression 350(1 - 0.3)
Which statement is true?
A)Each hour, the population decreases by 3%
B)Each hour, the population increase by 3 bacteria.
c) The hourly decay rate for the population is 30%
D)The initial population of bacteria is 245.
Answer:
D
Step-by-step explanation:
An altitude is drawn from the vertex of an isosceles triangle, forming a right angle and two congruent triangles. As a result, the altitude cuts the base into two equal segments. The length of the altitude is 9 inches, and the length of the base is 7 inches. Find the triangle’s perimeter. Round to the nearest tenth of an inch. Thanks!!!
Answer:
26.3 inchesStep-by-step explanation:
We separate into 2 congruent triangles. Both have a base of 3.5 and a height of 9 inches. Using pythagoran theorum, we will square 9 and 3.5
81 and 12.25
Add them up and it's 93.25
[tex]\sqrt{93.25}[/tex]=9.65660395791
Now since it's isosceles, the other side will also be 9.65660395791.
9.65660395791 + 9.65660395791 + 7 =26.3132079158
You want it rounded to the nearest tenth, so 26.3
Please help
Technology required. Function fis defined by f(x) = 3x + 5 and function g
is defined by g(x) = (1.1)x.
1. Complete the table with values of f(x) and g(x). When necessary, round to 2 decimal places.
2. Which function do you think grows faster? Explain your reasoning.
3. Use technology to create graphs representing fand g. What graphing window do you have to use to see the value of x where g becomes greater than f for that x?
The Complete value for the Table is
x f(x) g(x)
1 8 1.1
5 20 1.61051
10 35 2.5937
20 65 6.7274
The function f(x) grows with faster rate.
What is Function?A function is an expression, rule, or law in mathematics that describes a relationship between one variable (the independent variable) and another variable (the dependent variable).
Given:
We have f(x) = 3x+ 5 and g(x) = [tex](1.1)^x[/tex]
The Complete value for the Table is
x f(x) g(x)
1 3x+ 5 = 3(1) +5= 8 1.1
5 3x+ 5 = 3(5) +5= 20 [tex](1.1)^5[/tex]= 1.61051
10 3x+ 5 = 3(10) +5= 35 [tex](1.1)^{10}[/tex]= 2.5937
20 3x+ 5 = 3(20) +5= 65 [tex](1.1)^{20}[/tex]= 6.7274
So, the function f(x) grows with faster rate.
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Tom used a photocopier to dilate the design for a monorail track system. The figure shows the design and its photocopy: Two irregular quadrilaterals ABCD and EFGH are drawn. EFGH is the photocopy of design ABCD. AB measures 17 meters, AD measures 7 meters. No dimensions are shown on EFGH. The ratio of BC:FG is 1:2. What is the length, in meters, of side EH on the photocopied image? 7 14 17 34
In the congruent geometry , length of EH is 14 meters .
What is congruent geometry?
In a congruent geometry, the shapes that are so identical. can be superimposed on themselves.
Two objects that are identical in terms of their dimensions and shape are said to be congruent in geometry. Further, two shapes are said to be congruent to one another if they can be turned, flipped, or moved in the same direction.
As a result, two congruent figures that have been drawn on a piece of paper can be cut out and positioned over one another to perfectly match.
Here, The figure shows the design and its photocopy. The ratio of BC: FG is 1:2.
Since the two figures are congruent
BC/FG= AD/EH
1/2 = 7/EH (from figure AD = 7 )
EH = 2*7
EH = 14 m
Thus, the required length of EH is 14 meters.
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Let N E N be such that N > 2. Let Ω be the set of non-empty subsets of {1,...,N}, i.e. Ω = {ω C {1,...,N}: ω ≠0}. Let F be the o-algebra on Ω formed by all subsets of Ω and P be the uniform probability measure on (Ω, F). For ω E Ω, let X and Y be the random variables defined as X(ω) = max(ω) and Y(ω) = min(ω) So that, for example, if w = {1, 2, N} then X(W)= N and Y (W) = 1. (a) Show that the probability mass function of X is px (n) = 2^n-1 / 2^N – 1 n € {1,...,N} and 0 otherwise. (b) For any t € R, compute the value of the function 0: R -> R defined as ɸ(t) = E [2^tx] [2 marks] (c) Show that the joint probability mass function of (X,Y) is 1 / 2^N -1, for m
Pxx^(n, m) = { 2^(n-m-1) / 2^N -1 for n=m, n, m{1,...,N}
0, otherwise [2 marks] (d) Determine the probability mass function of W - X - Y. [3 marks)
(a) The probability that X = n is 2^n-1 / 2^N - 1.
(b) ɸ(t) = E[2^tx] = ∑_{n=1}^N 2^tx P(X=n) = ∑_{n=1}^N 2^tx (2^n-1 / 2^N - 1) = (2^t / 2^N - 1) ∑_{n=1}^N 2^(n-1)t = (2^t / 2^N - 1) (2^Nt - 1) / (2^t - 1) = 2^Nt / (2^N - 1)
(c) The probability that X = n and Y = m is 2^(n-m-1) / 2^N - 1.
(d) This is the same as finding the probability that the difference between the maximum and minimum elements of a subset of {1,...,N} is k. If k = 0, then the only subsets with maximum and minimum elements differing by k are the single-element subsets, so P(W=k) = N / 2^N - 1. If k > 0, then there are (N-k) choices for the minimum element m and 2^(k-1) subsets of {m+1,...,m+k-1}, so P(W=k) = (N-k) 2^(k-1) / 2^N - 1.
To find the probability mass function of X, we need to find the probability that X = n for each n ∈ {1,...,N}. This is the same as finding the probability that the maximum element of a subset of {1,...,N} is n. There are 2^n-1 subsets of {1,...,n-1}, and each of these subsets can be combined with n to form a subset of {1,...,N} with maximum element n. Therefore, the probability that X = n is 2^n-1 / 2^N - 1.
To find the value of ɸ(t) for any t ∈ R, we need to compute the expected value of 2^tx. Using the formula for expected value, we get:
ɸ(t) = E[2^tx] = ∑_{n=1}^N 2^tx P(X=n) = ∑_{n=1}^N 2^tx (2^n-1 / 2^N - 1) = (2^t / 2^N - 1) ∑_{n=1}^N 2^(n-1)t = (2^t / 2^N - 1) (2^Nt - 1) / (2^t - 1) = 2^Nt / (2^N - 1)
To find the joint probability mass function of (X,Y), we need to find the probability that X = n and Y = m for each n,m ∈ {1,...,N}. If n = m, then the only subset of {1,...,N} with maximum element n and minimum element m is {n}, so P(X=n, Y=m) = 1 / 2^N - 1. If n ≠ m, then there are 2^(n-m-1) subsets of {m+1,...,n-1}, and each of these subsets can be combined with m and n to form a subset of {1,...,N} with maximum element n and minimum element m. Therefore, the probability that X = n and Y = m is 2^(n-m-1) / 2^N - 1.
To find the probability mass function of W = X - Y, we need to find the probability that W = k for each k ∈ {0,...,N-1}. This is the same as finding the probability that the difference between the maximum and minimum elements of a subset of {1,...,N} is k. If k = 0, then the only subsets with maximum and minimum elements differing by k are the single-element subsets, so P(W=k) = N / 2^N - 1. If k > 0, then there are (N-k) choices for the minimum element m and 2^(k-1) subsets of {m+1,...,m+k-1}, so P(W=k) = (N-k) 2^(k-1) / 2^N - 1.
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Use an appropriate infinite series method about x = 0 to find two solutions of the given differential equation. (Enter the first four nonzero terms for each linearly independent solution, if there are fewer than four nonzero terms then enter all terms. In each case, the first term has been provided for you. )
3xy" y+y=0
Y₁ = 1 -
12 = x2/3
The two linearly independent solutions are y₁(x) = 1 - (x²/3) + (x⁴/81) - (x⁶/2187) + … and y₂(x) = x - (x³/9) + (x⁵/405) - (x⁷/21870) + ...
We can use the power series method to find the two solutions of the given differential equation:
Let y = ∑anxn be a power series solution. Then, we have:
y' = ∑nanxn-1
y'' = ∑nan(n-1)xn-2
Substituting these into the differential equation and equating coefficients of like powers of x, we get:
3x∑nan(n-1)xn-1 + ∑anxn = 0
Simplifying, we obtain:
∑[3nan(n-1) + an-1]xn = 0
Since x ≠ 0, the coefficients of like powers of x must be zero. This gives us the recurrence relation:
an = -an-1/(3n(n-1)), n ≥ 1
Using this recurrence relation and the initial condition y(0) = 1, we can find the power series solution:
y₁(x) = 1 - (x²/3) + (x⁴/81) - (x⁶/2187) + ...
To find a second linearly independent solution, we can use the method of reduction of order. Let y₂(x) = v(x)y₁(x). Then, we have:
y₂' = v'y₁ + vy₁'
y₂'' = v''y₁ + 2v'y₁' + vy₁''
Substituting these into the differential equation and simplifying, we get:
v''y1x + (2v'y1' + vy1'')x + 3vy1 = 0
Since y₁(x) is a solution, we have:
y₁'' + (1/3x)y₁ = 0
Multiplying by y1 and integrating, we obtain:
(y₁')² + y₁²/3 = C
where C is a constant of integration. Using the initial condition y₂(0) = 0, we can find the second solution:
y₂(x) = x - (x³/9) + (x⁵/405) - (x⁷/21870) + ...
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I need help with this please
The amount of Kingston's birth weight that was much bigger than Karmichael would be = 1 pound 4 ounces
How to calculate the birth weight of Kingston?The weight of Kingston at birth = 9 pounds 2 ounces
The weight of Karmichael at birth = 7 pounds 11 ounces.
Next, the weights are converted to ounces such as:
1 pound = 16 ounces
For 9 pounds = 16×9 = 144 ounces+ 2 ounces = 146 ounces
For 7 pounds = 16×7 = 112 ounce + 11 ounces = 123 ounce
The difference between the birth weight of Kingston and Karmichael = 146-123 = 23 ounce
Convert 23 ounce to pounds = 23/16 = 1 pound 4 ounces
Therefore, amount of Kingston's birth weight that was much bigger than Karmichael would be = 1 pound 4 ounces.
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What is the volume of a sphere with a diameter of 8.6 m, rounded to the nearest
tenth of a cubic meter?
Answer:
The volume of a sphere with a diameter of 8.6 m = 333.0 m^3
Step-by-step explanation:
Solving a compound linear inec Solve the compound inequality. 4v+3<=27 or ,2v-4>=0
The solution of the inequaltiy is v <= 6 or v >= 2.
To solve the compound inequality 4v+3<=27 or 2v-4>=0, we need to solve each inequality separately and then combine the solutions.
For the first inequality, 4v+3<=27, we can isolate the variable on one side of the inequality by subtracting 3 from both sides:
4v <= 24
Next, we can divide both sides by 4 to get:
v <= 6
For the second inequality, 2v-4>=0, we can isolate the variable on one side of the inequality by adding 4 to both sides:
2v >= 4
Next, we can divide both sides by 2 to get:
v >= 2
Now, we can combine the solutions to get the final solution for the compound inequality:
v <= 6 or v >= 2
This means that the solution set includes all values of v that are less than or equal to 6 or greater than or equal to 2.
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Researchers studied a random sample of high school students who participated in interscholastic athletics to learn about the risk of lower-extremity injuries (anywhere between hip and toe) for interscholastic athletes. Of 997 participants in girls' soccer, 79 experienced lower-extremity injuries. Of 1,664 participants in boys' soccer, 156 experienced lower-extremity injuries. (a) Write null and alternative hypotheses about sex and the risk of a lower-extremity Injury while playing interscholastic soccer. Null hypothesis: Sex and risk of a lower-extremity injury in interscholastic soccer are related variables. Alternative hypothesis: Sex and risk of a lower-extremity Injury in interscholastic soccer are not related variables. Null hypothesis: Sex and risk of a lower extremity injury in interscholastic soccer are explanatory variables. Alternative hypothesis: Sex and risk of a lower-extremity injury in interscholastic soccer are not explanatory variables. Null hypothesis: Sex and risk of a lower extremity injury in interscholastic soccer are not explanatory variables. Alternative hypothesis: Sex and risk of a lower-extremity injury in interscholastic soccer are explanatory variables. Null hypothesis: There is a weak relationship between sex and risk of a lower-extremity injury in interscholastic soccer. Alternative hypothesis: There is a strong relationship between sex and risk of a lower-extremity injury in interscholastic soccer Null hypothesis: Sex and risk of a lower-extremity injury in interscholastic soccer are not related variables. Alternative hypothesis: Sex and risk of a lower-extremity injury in interscholastic soccer are related variables. (b) For these data, the value of the chi-square statistic is 1. 63, and the p-value for the chi-square test is 0. 202. Based on these results, state a conclusion about the two variables in this situation and explain how you came to this conclusion. The p-value is greater than the 1. 63 X standard for significance, so there is not sufficient evidence to be able to conclude that the variables are related. (c) For each sex separately, calculate the percent of participants who had a lower-extremity injury. (Round your answers to one decimal place. ) girls 7. 9 % boys 9. 4 % Explain how the difference between these percentages is consistent with the conclusion you stated in part (b). These percentages are fairly similar , which suggests that sex and risk of a lower-extremity injury in interscholastic soccer are not related
a) Null hypothesis: Sex and risk of a lower-extremity injury in interscholastic soccer are not related variables. Alternative hypothesis: Sex and risk of a lower-extremity injury in interscholastic soccer are related variables. So the option E is correct.
b) There is insufficient information to draw the conclusion that the variables are connected even though the value is greater than the 0.05 threshold for significance.
c) The difference between these percentage are fairly similar which suggests that sex and risk of lower-extremity Injury in interscholastic soccer are not related.
a) The null and alternative hypotheses about sex and the risk of a lower-extremity Injury while playing interscholastic soccer is:
Null hypothesis: Sex and lower-extremity injury risk in interscholastic soccer are unrelated factors. Alternative hypothesis: sex and the likelihood of sustaining a lower-extremity injury in collegiate soccer are associated variables.
So the option E is correct.
b) The conclusion is that there is no significant relationship between the gender of the athlete and the risk of lower-extremity injuries. This conclusion is based on the fact that the chi-square statistic (1.63) and the corresponding p-value (0.202) are both greater than the 1.63 X standard for significance. This indicates that there is not enough evidence to show that the variables are related.
c) Of 997 participants in girls' soccer, 79 experienced lower-extremity injuries. Of 1,664 participants in boys' soccer, 156 experienced lower-extremity injuries.
Girls = 79/997 × 100 = 7.92%
Boys = 156/1664 × 100 = 9.38%
The difference between the two percentages is consistent with the conclusion that there is no significant relationship between the gender of the athlete and the risk of lower-extremity injuries because the difference between the two percentages is small.
This indicates that the risk of lower-extremity injuries is relatively similar for both boys and girls, which is consistent with the conclusion that there is no significant relationship between gender and the risk of lower-extremity injuries.
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The complete question is:
Researchers studied a random sample of high school students who participated in interscholastic athletics to learn about the risk of lower-extremity injuries (anywhere between hip and toe) for interscholastic athletes. Of 997 participants in girls' soccer, 79 experienced lower-extremity injuries. Of 1,664 participants in boys' soccer, 156 experienced lower-extremity injuries.
(a) Write null and alternative hypotheses about sex and the risk of a lower-extremity Injury while playing interscholastic soccer.
A. Null hypothesis: Sex and risk of a lower-extremity injury in interscholastic soccer are related variables. Alternative hypothesis: Sex and risk of a lower-extremity Injury in interscholastic soccer are not related variables.
B. Null hypothesis: Sex and risk of a lower extremity injury in interscholastic soccer are explanatory variables. Alternative hypothesis: Sex and risk of a lower-extremity injury in interscholastic soccer are not explanatory variables.
C. Null hypothesis: Sex and risk of a lower extremity injury in interscholastic soccer are not explanatory variables. Alternative hypothesis: Sex and risk of a lower-extremity injury in interscholastic soccer are explanatory variables.
D. Null hypothesis: There is a weak relationship between sex and risk of a lower-extremity injury in interscholastic soccer. Alternative hypothesis: There is a strong relationship between sex and risk of a lower-extremity injury in interscholastic soccer
E. Null hypothesis: Sex and risk of a lower-extremity injury in interscholastic soccer are not related variables. Alternative hypothesis: Sex and risk of a lower-extremity injury in interscholastic soccer are related variables.
(b) For these data, the value of the chi-square statistic is 1. 63, and the p-value for the chi-square test is 0. 202. Based on these results, state a conclusion about the two variables in this situation and explain how you came to this conclusion. The p-value is greater than the 1. 63 X standard for significance, so there is not sufficient evidence to be able to conclude that the variables are related.
(c) For each sex separately, calculate the percent of participants who had a lower-extremity injury. (Round your answers to one decimal place. ) girls 7.9% boys 9.4%. Explain how the difference between these percentages is consistent with the conclusion you stated in part (b). These percentages are fairly similar , which suggests that sex and risk of a lower-extremity injury in interscholastic soccer are not related
1 and 2 are complementary. if 1 is x + 4 and 2 is 23 + 2 x, find the measure of 1 and 2
Answer:
Step-by-step explanation:
x + 4 + 2x + 23 = 90
3x + 27 = 90
3x = 63
x = 21
x + 4 = 21 + 4 = 25 for <1
2(21) + 23 = 42 + 23 = 65 for <2
Prove for the simple decision sapling (sure thing of value r vs. risky gamble), that EVPI > 0. Also prove that if the random payoff of the gamble, call it G, is replaced by the random variable G+Y where Y is independent of G and EY = 0 (so G’ = G+Y is a noisy, or more variable, version of the original gamble G), then the EVPI will be larger. a
To prove that EVPI > 0 for the simple decision sapling, we need to understand what EVPI is. EVPI stands for Expected Value of Perfect Information, and it is the difference between the expected value of the decision with perfect information and the expected value of the decision without perfect information.
For the simple decision sapling, we have two options: a sure thing of value r, and a risky gamble with a random payoff G. The expected value of the decision without perfect information is simply the maximum of the two options:
EV = max(r, EG)
The expected value of the decision with perfect information is the maximum of the two options, knowing the outcome of the gamble:
EVPI = max(r, G) - max(r, EG)
Since we know that G is a random variable, the expected value of G is simply the average of all possible outcomes. Therefore, the expected value of the decision with perfect information is simply the maximum of the two options, knowing the average outcome of the gamble:
EVPI = max(r, EG) - max(r, EG) = 0
Therefore, EVPI > 0 for the simple decision sapling.
To prove that the EVPI will be larger if the random payoff of the gamble is replaced by a noisy version of the original gamble, we need to understand how the expected value of the decision changes with the addition of the noise variable Y. The expected value of the decision without perfect information is now:
EV = max(r, EG + EY) = max(r, EG)
Since EY = 0, the expected value of the decision without perfect information does not change. However, the expected value of the decision with perfect information now becomes:
EVPI = max(r, G + Y) - max(r, EG)
Since Y is a random variable with mean 0, the expected value of Y is 0. However, the variance of Y is not necessarily 0, which means that the addition of Y adds variability to the decision. This means that the expected value of the decision with perfect information is now larger, because we have more information about the possible outcomes of the gamble. Therefore, the EVPI will be larger when the random payoff of the gamble is replaced by a noisy version of the original gamble.
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Using the rational root theorem, list out all possibl f(x)=-x+14x^(3)-18x^(2)-4x^(5)-26x^(4)+8
The possible rational roots of f(x) are ±1, ±2, ±4, ±8, ±1/2, and ±1/4.
The Rational Root Theorem states that if a polynomial [tex]f(x) = anxn + an-1xn-1 + ... + a1x + a0[/tex] has a rational root, then it must be of the form p/q, where p is a factor of the constant term a0 and q is a factor of the leading coefficient an.
For the given polynomial [tex]f(x) = -x + 14x3 - 18x2 - 4x5 - 26x4 + 8[/tex], the constant term is 8 and the leading coefficient is -4.
The factors of 8 are ±1, ±2, ±4, and ±8. The factors of -4 are ±1, ±2, and ±4.
Therefore, the possible rational roots of f(x) are:
p/q = ±1/1, ±2/1, ±4/1, ±8/1, ±1/2, ±2/2, ±4/2, ±8/2, ±1/4, ±2/4, ±4/4, ±8/4
Simplifying gives us the possible rational roots:
±1, ±2, ±4, ±8, ±1/2, ±4/2, ±8/2, ±1/4, ±2/4, ±8/4
Simplifying further gives us the final list of possible rational roots:
±1, ±2, ±4, ±8, ±1/2, ±2, ±4, ±1/4, ±1/2, ±2
Removing duplicates, the final list of possible rational roots is:
±1, ±2, ±4, ±8, ±1/2, ±1/4
Therefore, the possible rational roots of f(x) are ±1, ±2, ±4, ±8, ±1/2, and ±1/4.
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The sum of two numbers is 42. The difference of the two numbers is 30. What are the two numbers. Let x be the larger number and y be the smaller number. Write an equation that expresses the information in the sentence 'The sum of two numbers is 42." _____________
Write an equation that expresses the information in the sentence 'The difference of the two numbers is 30." __________
Solve the system you have written above. The larger number, x is ______ The smaller number, y is _____
Answer:
x=36
y=6
system of equations
please someone complete this ive never struggled so bad
Answer:
Step-by-step explanation:
Im not really sure what the heck you doing because im only in 8th grade but right now im in central and i'm kind of good at math but i'll try. Lets see........ I think they would all just be $11.50, but like i said im only in 8th grade and not that good at this kind of math so before you answer it please check with someone else to see if i got it right for you, cause i don't want to give you the wrong answer and you fail it.
I’m class A, 28 out of 60 students are girls and in class B 44 out of the 60 students are girls. Which class has a higher percentage of girls
PLS HELPP
Class B has higher percentage of girls than class A that is 26.6 % more.
What is percentage?A percentage is a number or ratio that can be expressed as a fraction of 100. If you need to calculate the percentage of a number, divide the number by the whole number and multiply by 100. Percentages therefore mean 1 in 100.
Given,
Total number of students in class A = 60
Number of girls student in class A = 28
percentage of girls in class A
= (28/60)×100
= 0.467 × 100
= 46.7%
Total number of students in class B = 60
Number of girls student in class B = 44
percentage of girls in class A
= (44/60)×100
= 0.733 × 100
= 73.3 %
Difference in percentage of girls in class A and class B.
= 73.3% - 46.7%
= 26.6%
Hence, class B has 26.6 percent more girls than class A.
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Pls help me it’s to go from lease value to the greatest and closest to zero!
According to the information, the expressions that would match the descriptions would be both expressions have the same result, so it doesn't matter which description they are associated with. both are the largest and the smallest value.
How to find the correct expressions for the descriptions?To find the correct descriptions that match the expressions we must look at the graph. In this case q and n represent two numbers on the number line. In this case, to relate them to a description we must find the number to which they refer:
q = -1n = -4The expression q - n would be equal to -1 - 4 = -5The expression n - q would be equal to - 4 - 1 = -5In accordance with the above, we could say that the descriptions would look like this:
Closest to zero = qLeast value = q - nGreatest value = n - qIn this case, both expressions have the same result, so it doesn't matter which description they are associated with. both are the largest and the smallest value.
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PLSSSS HELP IF YOU TURLY KNOW THISSSS
Answer: 1
Step-by-step explanation:
3(x-2) = 4x+2
3x-6 = 4x+2
To move 3x, deduct it on both sides of the equation.
3x - 6 - (3x) = 4x + 2 - (3x)
0 - 6 = [1]x +2
Given:-
[tex] \tt \: 3( x - 2 ) = 4x + 2[/tex][tex] \: [/tex]
Solution:-
[tex] \tt \: 3( x - 2 ) = 4x + 2[/tex][tex] \: [/tex]
[tex] \tt \: 3x - 6 = 4x + 2[/tex][tex] \: [/tex]
[tex] \tt \: -6 = 4x - 3x + 2[/tex][tex] \: [/tex]
[tex] \tt \: -6 = 1x + 2 [/tex][tex] \: [/tex]
[tex] \tt \: -6 = 3x[/tex][tex] \: [/tex]
[tex] \tt \: \cancel\frac{ - 6}{3} = x[/tex][tex] \: [/tex]
[tex] \boxed{\tt \green{- 2 = x}}[/tex][tex] \: [/tex]
━━━━━━━━━━━━━━━━━━━━━━━
hope it helps ⸙
9. Verify that each equation is an identity. a. sin 2x = 2 tan x / 1 + tan^2 x b. tan x + cot x = 2 csc 2x
a) Sin 2x = 2 tan x / 1 + tan^2 x is an identity.
b) Tan x + cot x = 2 csc 2x is an identity.
To verify that each equation is an identity, we will simplify both sides of the equation and show that they are equal.
For part a, we will use the double angle formula for sine and the Pythagorean identity.
sin 2x = 2 sin x cos x
2 tan x / 1 + tan^2 x = 2 sin x / cos x / 1 + sin^2 x / cos^2 x
= 2 sin x / cos x / cos^2 x / cos^2 x
= 2 sin x / cos x / 1 - sin^2 x
= 2 sin x / cos x / cos^2 x
= 2 sin x cos x
Therefore, sin 2x = 2 tan x / 1 + tan^2 x is an identity.
For part b, we will use the definitions of the trigonometric functions and the double angle formula for cosecant.
tan x + cot x = sin x / cos x + cos x / sin x
= (sin^2 x + cos^2 x) / (sin x cos x)
= 1 / (sin x cos x)
= 2 / (2 sin x cos x)
= 2 / sin 2x
= 2 csc 2x
Therefore, tan x + cot x = 2 csc 2x is an identity.
In conclusion, we have verified that both equations are identities.
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Which of the following is the graph of the piecewise function f of x is equal to the piecewise function of the quantity square root of negative x minus 1 if x is less than or equal to negative 1 and the function negative x over the quantity x squared minus x minus 2 end quantity if negative 1 is less than x is less than 2 and the function log in base 2 of the quantity x minus 1 end quantity if x is greater than or equal to 2 question mark
The graph of the piece-wise function is given by the image presented at the end of the answer.
What is a piece-wise function?A piece-wise function is a mathematical function that is defined by different formulas or expressions over different intervals or pieces of its domain, that is, the function has different definitions based on it's input.
The intervals for which the function has different definitions are given as follows:
x ≤ -1.-1 < x < 2.x ≥ 2.The definitions for each interval are given as follows:
x ≤ -1 -> f(x) = sqrt(x - 1). -> outside domain of square root, hence does not show up on graph-1 < x < 2 -> f(x) = -x/(x² - x - 2).x ≥ 2. -> f(x) = log2(x - 1).Hence the graph of the function, containing these three definitions, is given by the image presented at the end of the answer.
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A city had a declining population from 1992 to 1998. The population in 1992 was 200,000. Each year for 6 years, the population declined by 3%. Write an exponential decay model to represent this situation.
The exponential decay model for this situation is P(t) = 200,000 * (1 - 0.03)^t
How to find the percentage from the total value?Suppose the value of which a thing is expressed in percentage is "a'
Suppose the percent that considered thing is of "a" is b%
Then since percent shows per 100 (since cent means 100), thus we will first divide the whole part in 100 parts and then we multiply it with b so that we collect b items per 100 items(that is exactly what b per cent means).
We need to Write the percent as a fraction in simplest form
16.24%
So, we can rewrite it as;
16.24 / 100
16 and 24/100 or 16 6/25
So the percent as a fraction is 16 6/25
We are given that;
Population in 1992=200000
Time=6years
Rate=3%
To write an exponential decay model to represent this situation, we can use the formula:
P(t) = P * (1 - r)ᵗ
where P(t) is the population after t years, P is the initial population, r is the annual rate of decline as a decimal, and t is the number of years. In this case, the initial population P is 200,000, the annual rate of decline r is 0.03 (since the population declines by 3% each year), and t is the number of years from 1992, so t = 0 corresponds to 1992 and t = 6 corresponds to 1998.
P(t) = 200,000 * (1 - 0.03)^t
where t is the number of years from 1992 to 1998.
Therefore, by the given percent answer will be P(t) = 200,000 x (1 - 0.03)^t
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Perform the indicated operation on the algebraic expressions. Sim (u-v)(u^(2)+uv+v^(2))
The simplified expression is u^(3) - v^(3).
To perform the indicated operation on the algebraic expressions, we need to multiply each term in the first expression by each term in the second expression and then simplify the resulting expression.
Step 1: Multiply each term in the first expression by each term in the second expression:
(u)(u^(2)) + (u)(uv) + (u)(v^(2)) - (v)(u^(2)) - (v)(uv) - (v)(v^(2))
Step 2: Simplify the resulting expression by combining like terms:
u^(3) + u^(2)v + uv^(2) - u^(2)v - uv^(2) - v^(3)
Step 3: Simplify further by canceling out terms that are equal but opposite in sign:
u^(3) - v^(3)
Therefore, the simplified expression is u^(3) - v^(3).
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Mengxi has $10 000 to invest. She invests part in a term deposit paying 5% /year, and the remainder in Canada Savings Bonds paying 3. 5% /year. At the end of the year, she has earned simple interest of $413. How much did she invest at each rate? (Algebra)
Mengxi invested $4,200 in the term deposit paying 5% /year and $5,800 in Canada Savings Bonds paying 3.5% /year. The total interest earned is $413.
Let's assume that Mengxi invests x dollars in the term deposit paying 5% /year, and the remaining (10000 - x) dollars in Canada Savings Bonds paying 3.5% /year.
At the end of the year, Mengxi earns a total of $413 in simple interest. The interest earned from the investment in the term deposit is calculated as 0.05x, while the interest earned from the investment in Canada Savings Bonds is 0.035(10000 - x).
Thus, we can write the following equation to represent the total interest earned:
0.05x + 0.035(10000 - x) = 413
Simplifying the equation, we get:
0.015x + 350 = 413
0.015x = 63
x = 4200
Therefore, Mengxi invested $4,200 in the term deposit paying 5% /year, and $5,800 (10000 - 4200) in Canada Savings Bonds paying 3.5% /year.
To check the answer, we can calculate the interest earned from each investment and add them up:
0.05(4200) + 0.035(5800) = 210 + 203 = 413
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