Hello please help asap!! i will mark brainliest and this is worth 20 points!!!!! tysm
Answers
The least number of colors needed to correctly color in the sections of the given picture so that no two touching sections have the same color is 5.
Let's say we begin with a single triangle. This triangle will be the first of a series of seven subsequent triangles, none of which will have any portions contact. If a square were the initial shape, then there are five shapes—three of which are squares and the other two are triangles—that don't have any touching portions. Given that a square takes up twice as much room, this value is probably lower, but location also matters. Therefore, for the given figure, the least number of colored areas you can color in that yet meet the required criteria is 5.
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Related Questions
Will mark brainliest
Answers
Using the given definition and [tex]\Delta x=\frac{0-(-2)}n=\frac2n[/tex], we have
[tex]\displaystyle \int_{-2}^0 (7x^2+7x) \,dx = \lim_{n\to\infty} \sum_{i=1}^n \left(7\left(-2+\frac{2i}n\right)^2 + 7\left(-2+\frac{2i}n\right)\right) \frac2n \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac2n \sum_{i=1}^n \left(14 - \frac{42i}n + \frac{28i^2}{n^2}\right)[/tex]
Recall the well-known power sum formulas,
[tex]\displaystyle \sum_{i=1}^n 1 = \underbrace{1 + 1 + 1 + \cdots + 1}_{n\,\rm times} = n[/tex]
[tex]\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2[/tex]
[tex]\displaystyle \sum_{i=1}^n i^2 = 1 + 4 + 9 + \cdots + n^2 = \frac{n(n+1)(2n+1)}6[/tex]
Reducing our sum leads to
[tex]\displaystyle \int_{-2}^0 (7x^2+7x) \,dx = \lim_{n\to\infty} \frac2n \left(\frac{7n}3 - 7 + \frac{14}{3n}\right) = \lim_{n\to\infty} \left(\frac{14}3 - \frac{14}n + \frac{28}{3n^2}\right)[/tex]
As [tex]n[/tex] goes to ∞, the rational terms containing [tex]n[/tex] will converge to 0, and the definite integral converges to
[tex]\displaystyle \int_{-2}^0 (7x^2+7x) \,dx = \boxed{\frac{14}3}[/tex]
Two factory plants are making tv panels yesterday plant A produced 3000 panels. Ten percent of the panels from plant A were defective and Three percent from plant B were defective. How many panels did plant be produce if the overall percentage of defective panels from the two plants was 6%. Thanks for the help
Answers
Answer: Plant B produced 4000 panels
Step-by-step explanation: Let the number of panels produced by plant A = a and the number of panels produced by plant B = x. In total, we have 3000 + x. Now for the defective panels. 10% of plant a + 3% of plant b = 6% of both a and b. 300 + 3/100(x) = 6/100(3000 + x). 300 + 0.03x = 180 + 0.06x. Now, we simplify. 300 - 180 = 0.06x = 0.03x. So, 120 = 0.03x and 40 = 1% of x. This means x = 4000. So, therefore, Plant B produced 4000 panels. We can check our work by using the value of x we got. 4000 + 3000 = 7000. 6% of 7000 = 420. 120 = 3% of 4000. 120 + 300 = 420 so the answer we got is correct. Hope this helped
What is the best estimated volume of the figure?
can anyone solve this for me with a detailed explanation step by steps and reasoning so i can grasp concepts
Answers
The answer is 756 m³.
The formula to find volume is : length × width × height
To estimate, round each dimension to the nearest whole number.
⇒ 12.3 ≅ 12
⇒ 7.2 ≅ 7
⇒ 8.6 ≅ 9
Volume = 12 × 7 × 9
Volume = 84 × 9
Volume = 756 m³
Answer:
The ans to this question is 756 which is Option (B)
Step-by-step explanation:
Estimate volume = rounding every digit
= 12.3 = 12
= 7.2 = 7
= 8.6 = 9
= 12x7x9 = 756
Write your answer as a fraction in simplest form.
-1 3/4 • 2 1/8
Answers
Which expression represents seven less than the product of thirteen times a number, and two squared?
(13x + 22) · 7
7(13x · 22)
7 + 13x + 22
7 + 13x · 22
(13x · 22) − 7
Answers
Seven less than the product of thirteen times a number, and two squared is as follows;
(13x × 2²) - 7
How to represent an expression?The expression says seven less than the product of thirteen times a number, and two squared.
Let
the number = x
Hence, the product of thirteen times a number, and two squared is as follows:
13(x)(2²)
Therefore, seven less than the product of thirteen times a number, and two squared is as follows;
(13x × 2²) - 7
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What are the first five terms in the recursive sequence defined by the following? (only one is correct)
a1= 1
a2=1
an= an-2+an-1
a) {1,1,2,3,5}
b) {1,1,0,-1,-1}
c) {2,3,5,8,13}
d) {1,-1,2,-3,5}
Answers
Answer:
d 1-12-3-5 is the answer
Sodas cost $1.75 each. Write a direct
proportional equation using (s) for sodas and
(t) for total cost.
Answers
Answer:
t = 1.75(s)
Step-by-step explanation:
I gotchu
pleaseee helpp
which of the following enequalities would produce the region indicated on the graph below
Answers
Answer: C
Step-by-step explanation:
The line [tex]y=x+2[/tex] is shaded below and is solid.
This eliminates all the options except for C.
Using two six-sided number cubes, each labeled with the numbers 1 through 6, event A is rolling a sum less than 6. Which of the following shows the sample space of event A?
{(1, 1), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3)}
{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
{(1, 1), (1, 2), (1, 3), (1, 5), (2, 1), (2, 2), (2, 3), (3, 1), (3, 3), (4, 1)}
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (3, 3), (4, 1), (4, 2)}
Answers
the sample space is:
{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1), (2, 2), (2, 3), (3, 2)}
Which of the following shows the sample space of event A?Event A is rolling a sum less than 6.
Let's define the possible elements in this experiment as:
(outcome of dice 1, outcome of dice 2)
The outcomes where the sum is less than 6 are:
dice 1 dice 2 sum
1 1 2
1 2 3
1 3 4
1 4 5
2 1 3
3 1 4
4 1 5
2 2 4
3 2 5
2 3 5
So there are 10 outcomes, then the sample space is:
{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1), (2, 2), (2, 3), (3, 2)}
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Answer:
B) {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
Step-by-step explanation:
If the question said less than 6 meaning you have to find all possible solution that are 5 or lower.
However, if the problem said equal or less than 6 then you have to find all possible solution that are 6 or lower.
B option is only option that don't have sum of 6. Therefore, option B is correct.
how many significant figures are in this number 6.01x10^3
Answers
Answer:
3
Step-by-step explanation:
6.01 has 3 sig fig
I think that's how I do it
A yo-yo is moving up and down a string so that its velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. The initial position of the yo-yo at time t = 0 is x = 3.
Part A: Find the average value of v(t) on the interval open bracket 0 comma pi over 2 close bracket. (10 points)
Part B: What is the displacement of the yo-yo from time t = 0 to time t = π? (10 points)
Part C: Find the total distance the yo-yo travels from time t = 0 to time t = π. (10 points)
Answers
Part A - The average value of v(t) over the interval (0, π/2) is 6/π
Part B - The displacement of the yo-yo from time t = 0 to time t = π is 0 m
Part C - The total distance the yo-yo travels from time t = 0 to time t = π is 6 m.
Part A: Find the average value of v(t) on the interval (0, π/2)The average value of a function f(t) over the interval (a,b) is
[tex]f(t)_{avg} = \frac{1}{b - a} \int\limits^b_a {f(t)} \, dx[/tex]
So, since velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. Its average value over the interval (0, π/2) is given by
[tex]v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {v(t)} \, dt[/tex]
Since v(t) = 3cost, we have
[tex]v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {3cos(t)} \, dt\\= \frac{3}{\frac{\pi }{2}} \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= \frac{6}{{\pi}} [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= \frac{6}{{\pi}} [{sin(\frac{\pi }{2})} - sin0]\\ = \frac{6}{{\pi}} [1 - 0]\\ = \frac{6}{{\pi}} [1]\\ = \frac{6}{{\pi}}[/tex]
So, the average value of v(t) over the interval (0, π/2) is 6/π
Part B: What is the displacement of the yo-yo from time t = 0 to time t = π?To find the displacement of the yo-yo, we need to find its position.
So, its position x = ∫v(t)dt
= ∫3cos(t)dt
= 3∫cos(t)dt
= 3sint + C
Given that at t = 0, x = 3. so
x = 3sint + C
3 = 3sin0 + C
3 = 0 + C
C = 3
So, x(t) = 3sint + 3
So, its displacement from time t = 0 to time t = π is
Δx = x(π) - x(0)
= 3sinπ + 3 - (3sin0 + 3)
= 3 × 0 + 3 - 0 - 3
= 0 + 3 - 3
= 0 + 0
= 0 m
So, the displacement of the yo-yo from time t = 0 to time t = π is 0 m
Part C: Find the total distance the yo-yo travels from time t = 0 to time t = π. (10 points)The total distance the yo-yo travels from time t = 0 to time t = π is given by
[tex]x(t) = \int\limits^{\pi}_0 {v(t)} \, dt\\= \int\limits^{\pi }_0 {3cos(t)} \, dt\\= 3 \int\limits^{\pi }_0 {cos(t)} \, dt\\ = 3 \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt + 3\int\limits^{\pi }_{\frac{\pi }{2}} {cos(t)} \, dt\\= 3 \times 2\int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= 6 [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= 6[{sin\frac{\pi }{2} - sin0]\\\\= 6[1 - 0]\\= 6(1)\\= 6[/tex]
So, the total distance the yo-yo travels from time t = 0 to time t = π is 6 m.
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Use the table values below to select the correct statement.
Answers
The true statement is (c) the function f(x) is an exponential function
How to determine the correct statement?From the table, we can see that:
As x increases by 2.y does not increase with the same common differenceThis means that the table represents an exponential function
The rate is then calculated as:
f(n) = f(1) * r^n-1
Using f(3), we have
102 = 17 * r^3-1
This gives
r^2 = 6
Take the square roots
r = 2.45
Hence, the true statement is (c) the function f(x) is an exponential function
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What are the potential zeros of f(x)=6x^4+ 2x^3 - 4x^2 +2?
Answers
The potential zeros of f(x)=6x^4+ 2x^3 - 4x^2 +2 are ±(1, 1/2, 1/3, 1/6, 2, 2/3)
How to determine the potential zeros of the function f(x)?The function is given as:
f(x)=6x^4+ 2x^3 - 4x^2 +2
For a function P(x) such that
P(x) = ax^n +...... + b
The rational roots of the function p(x) are
Rational roots = ± Possible factors of b/Possible factors of a
In the function f(x), we have:
a = 6
b = 2
The factors of 6 and 2 are
a = 1, 2, 3 and 6
b = 1 and 2
So, we have:
Rational roots = ±(1, 2)/(1, 2, 3, 6)
Split the expression
Rational roots = ±1/(1, 2, 3, 6)/ and ±2/(1, 2, 3, 6)
Evaluate the quotient
Rational roots = ±(1, 1/2, 1/3, 1/6, 2, 1, 2/3, 1/3)
Remove the repetition
Rational roots = ±(1, 1/2, 1/3, 1/6, 2, 2/3)
Hence, the potential zeros of f(x)=6x^4+ 2x^3 - 4x^2 +2 are ±(1, 1/2, 1/3, 1/6, 2, 2/3)
The complete parameters are:
The function is given as:
f(x) = 3x^3 + 2x^2 + 3x + 6
The potential zeros of f(x)=6x^4+ 2x^3 - 4x^2 +2 are ±(1, 1/2, 1/3, 1/6, 2, 2/3)
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What is the probability that the mean score for 10 randomly selected people who took the LSAT would be above 157? Round your answer to three decimal places. (Example: 0.398)
Answers
Using the normal distribution, there is a 0.007 = 0.7% probability that the mean score for 10 randomly selected people who took the LSAT would be above 157.
Normal Probability DistributionThe z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The z-score measures how many standard deviations the measure is above or below the mean. Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].Researching this problem on the internet, the parameters are given as follows:
[tex]\mu = 150, \sigma = 9, n = 10, s = \frac{9}{\sqrt{10}} = 2.85[/tex]
The probability is one subtracted by the p-value of Z when X = 157, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
Z = (157 - 150)/2.85
Z = 2.46
Z = 2.46 has a p-value of 0.993.
1 - 0.993 = 0.007.
0.007 = 0.7% probability that the mean score for 10 randomly selected people who took the LSAT would be above 157.
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How many three-digit positive integers have at least one prime digit?
Answers
Answer:
210
Step-by-step explanation:
[tex]512=2^9 \\[/tex] has 9 prime factors
Now, if you want to maximize the number of different prime factor, apply the following method: let [tex]p(n)[/tex] be the nth prime number, in increasing order. Let [tex]v[/tex] define as follow : [tex]v(0)=1,v(n+1)=v(n)p(n+1)[/tex] , and find n such that [tex]v(n) < 1000,v(n+1) > 1000[/tex]. n will be the number of factors and v(n) your integer.
[tex]v(1)=2v(0)=2v(2)=3v(1)=6v(3)=5v(2)=30v(4)=7v(3)=210v(5)=11v(4)=2310[/tex]
There you are, 210 , with only 4 different prime factors.
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Solve the inequality for x.
OA.
X S
5- 2/2 x ²
x 2
28
OB. x ≤ 7
OC.
28
9
OD. x ≥ 7
Answers
The solution to the inequality x-13<=7+4x is x >= -20/3
How to solve the inequality?The inequality expression is given as:
x-13<=7+4x
Add 13 to both sides of the inequality
x <= 20 + 4x
Subtract 4x from both sides of the inequality
-3x <= 20
Divide both sides of the inequality by -3
x >= -20/3
Hence, the solution to the inequality x-13<=7+4x is x >= -20/3
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Complete question
Solve the inequality for x
x-13<=7+4x
Alok started a business investing Rs 90, 000. After three
months Prabir joined him with capital of Rs 1,20000
If at the end of 2 years, the total profit made
by them was
Rs 96,000 what will the difference
between Alok and Prabir's share in it?
Answers
The difference between Alok and Prabir share exists 8000.
What will the difference between Alok and Prabir share?Given: Invested by Alok = Rs 90, 000
Invested by Prabir = Rs 1,20000
The time period of Alok = 3 months
The time period of Prabir = 2 years
They earn a profit = Rs 96,000.
Profit exists directly proportional to the product of the amount invested and the time period of investment.
8000 = 90000 [tex]*[/tex] 24/120000 [tex]*[/tex] 21
= 5/7
5x + 7x = 96000
x = 8000
first = 40000
second = 48000
so the difference exists at 8000.
The difference between Alok and Prabir share exists 8000.
Therefore, the correct answer is 8000.
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Fill in the P(X=x) values to give a legitimate probability distribution for the discrete random variable , whose possible values are 0, 1, 3, 4, and 5.
Answers
For a probability distribution to be represented, it is needed that P(X = 0) + P(X = 1) = 0.44. Hence one possible example is:
P(X = 0) = 0.40.P(X = 1) = 0.04.What is needed for a discrete random variable to represent a probability distribution?The sum of all the probabilities must be of 1, hence:
P(X = 0) + P(X = 1) + P(X = 3) + P(X = 4) + P(X = 5) = 1.
Then, considering the table:
P(X = 0) + P(X = 1) + 0.15 + 0.17 + 0.24 = 1
P(X = 0) + P(X = 1) + 0.56 = 1
P(X = 0) + P(X = 1) = 0.44.
Hence one possible example is:
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Can someone please please help me
Answers
Answer:
order: 2, 3, 1
Step-by-step explanation:
Reasonableness checks and your knowledge of integers and fractions will help you solve this. The offered questions are intended to help you think this through.
a.For an output of -31, the machine (x -2)² cannot possibly be last. Its output can only be positive.
machine 2, (x-2)², cannot be last
Also, machine 3 cannot be last. For the output to be -31, the input to machine 3 must be -1/31. Neither of the other machines can produce a fraction with the inputs they might receive.
b.
For an input of x=0, the machine 1/x cannot possibly be first. 1/0 is undefined.
The other two machines will give the following outputs for an input of 0:
machine 1: 4(0) -32 = -32
machine 2: (0 -2)² = 4
It is unlikely that machine 1 will be first, because the other two machines cannot do anything useful with -32 as an input.
machine 3, 1/x, cannot be first
c.The reasoning of part (a) tells you the last machine must be machine 1. The reasoning of part (b) tells you the first machine cannot be 3, so must be 2. The order of the machines must be ...
machine 2: (0 -2)² = 4 . . . . . . . using an input of 0machine 3: 1/4 = 1/4machine 1: 4(1/4) -32 = -31 . . . . desired outputI need help solving this problem!
Answers
Answer: A
Step-by-step explanation:
The domain is the range of possible x that doesn't make y impossible to find, in this case, all real numbers work
The range is the range of possible y that doesn't make x impossible to find in the inverse function, in this case, all real numbers work
The answer is A.
As any number can be replaced in place of x for the function y, the domain of the function is All Real Numbers. Since any number can be inputted, the result will also vary accordingly. Hence, the range of the function is All Real Numbers.
This is true for the listed function :
[tex]\boxed {y = \sqrt[3]{x-2} - 5}[/tex]
PLEASE HELP!
Let f be the function given by f (x) = (create an original sinusoidal function with an amplitude not equal to 1, a period not equal to 2π, and non-zero phase and vertical shifts).
ex: F of x equals negative one half times sine of quantity 3 times x plus pi over 2 end quantity minus 2
Part A: State the amplitude and vertical shift.
Part B: Determine the period of f (x), showing all necessary calculations.
Part C: Calculate the phase shift of the sinusoidal function with proper mathematical justification.
Part D: Graph the sinusoidal function by hand, using your answers from parts A–C.
Choose an angle θ, in radians, such that 2π < θ < 4π . Let θ = (create an original angle measure).
ex: Theta equals 13 pi over 6
Part E: Determine the exact value of cos θ using the sum formula. Show all necessary mathematical work.
Part F: Determine the exact value of sin θ using the difference formula. Show all necessary mathematical work.
Part G: Calculate the exact value of tan 2θ, using your answers from parts E – F.
Answers
The equation of the function f(x) is f(x) = 2 sin(π/2(x + 6)) - 3
How to create the sine function?A sine function is represented as:
f(x) = A sin(B(x + C)) + D
Where
A = Amplitude
Period = 2π/B
C = Phase shift
D = Vertical shift
The requirements in the question are:
Amplitude not equal to 1Period not equal to 2πNon-zero phase and vertical shiftsSo, we can use the following assumptions
A = 2
Period = 4
C = 6
D = -3
So, we have:
f(x) = 2 sin(B(x + 6)) - 3
The value of B is
4 = 2π/B
This gives
B = π/2
So, we have:
f(x) = 2 sin(π/2(x + 6)) - 3
The amplitude, vertical shift, period of f(x)and the phase shiftUsing the representations in (a), we have:
Amplitude = 2Vertical shift = -3Period = 4Phase shift = 6The graph of the functionSee attachment for the graph of f(x)
The value of cos θLet θ = 3π
So, we have:
cos(3π)
This is calculated as:
cos(3π) = cos(2π + π)
Expand
cos(3π) = cos(2π) *cos(π) - sin(2π) *sin(π)
Evaluate
cos(3π) = -1
The value of sin θLet θ = 3π
So, we have:
sin(3π)
This is calculated as:
sin(3π) = sin(2π + π)
Expand
sin(3π) = sin(2π) *cos(π) + cos(2π) *sin(π)
Evaluate
sin(3π) = 0
The value of tan 2θLet θ = 3π
So, we have:
tan(2 * 3π)
tan(6π)
This is calculated as:
tan(6π) = tan(3π + 3π)
Evaluate
tan(3π) = 0
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Write the ratio of 4 roses to 24 flowers
Answers
Initially we have 4:24 meaning 4 roses for 24 flowers. However, we can simplify this ration same as we do to fractions. If we divided both sides of the ration by 4, we get 1:6, or 1 rose for every 6 flowers.
Answer:
4:24 or also 1:6
Step-by-step explanation:
need heeeelp please
Answers
Answer:
(x, y) = (-√3/2, 1/2)
Step-by-step explanation:
The terminal point for that angle can be read from a unit circle chart.
On the attached chart, the point of interest is the first one above the -x axis on the left side. The chart tells you the coordinates are ...
(x, y) = (-√3/2, 1/2)
What is the sum of ∞Σn=1 2(1/5)^n-1? s=1/5 s=2/5 s=5/3 s=5/2
Answers
The sum of the sum notation ∞Σn=1 2(1/5)^n-1 is S= 5/2
How to determine the sum of the notation?The sum notation is given as:
∞Σn=1 2(1/5)^n-1
The above notation is a geometric sequence with the following parameters
Initial value, a = 2Common ratio, r = 1/5The sum is then calculated as
S = a/(1 - r)
The equation becomes
S = 2/(1 - 1/5)
Evaluate the difference
S = 2/(4/5)
Express the equation as products
S = 2 * 5/4
Solve the expression
S= 5/2
Hence, the sum of the sum notation ∞Σn=1 2(1/5)^n-1 is S= 5/2
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Answer:
S = 5/2
Thank me later
What's the Value of X in 1/2 + X = 5/6
Answers
Answer:
x = 1/3
Step-by-step explanation:
1/2 + X = 5/6
collecting the like terms
X = 5/6 - 1/2
making one fraction on the left side by taking the common denominator
X = (5 - 3)/6
X = 2/6
devide the numerator and denominator by 2, you get
X = 1/3
Madison finished 4/5 of her homework before dinner. What percent of Maddison’s homework is left to finish?
Answers
Answer:
1/5 of it
Step-by-step explanation:
5/5 represents all of her homework
5/5-4/5= 1/5
4/5 represents the homework she did before dinner
1/5 represents the homework left to do
What are the excluded values? –4 and –1 -1/2, 1, and 4 1 1 and 4
Answers
The excluded values are x = 4 and x - 1
How to determine the excluded values?The complete question is added as an attachment
The function is given as:
(2x^2 - 7x - 4)/(x^2 - 5x + 4)
Set the denominator to 0
x^2 - 5x + 4 = 0
Expand
x^2 - x - 4x + 4 = 0
Factorize the equation
x(x -1) - 4(x - 1) = 0
This gives
(x - 4)(x - 1) = 0
Solve for x
x = 4 and x - 1
Hence, the excluded values are x = 4 and x - 1
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Answer: Its D 1 and 4
Step-by-step explanation:
I took the assignment
|3x-1|=4 help pleaseeeeeeeeee
Answers
Answer: x = -1 or x = 5/3
Step-by-step explanation:
|3x - 1| = 4
=> 3x - 1 = 4 or 3x - 1 = -4
=>x = 5/3 or x = -1
x+2y=5 and 4x+12y=-20 elimination method
Answers
Answer:
x=25 and y=-10
Step-by-step explanation:
x+2y=5 ..........(1)
4x+12y=-20.........(2)
using elimination method.
multiply equ(1) by 4 and equ(2) by 1
so we have
x+2y=5..........*4
4x+12y=-20.........*1
4x+8y=20................(3)
4x+12y=-20.............(4)
subtract eq(4) from (3) we have
-4y=40
y=-10
substitute y=-10 in equation (1)
we have:
x+2(-10)=5
x-20=5
x=25
Help me with this equation, please. (Image Attached)
Answers
So, the equation is sin(x + y)/sin(x - y) = (1 + cotxstany)/(1 + cotxtany)
The question has to to with trigonometric identities?
What are trigonometric identities?Trigonometric identities are equations that show the relationship between the trigonometric ratios.
How to solve the equation?Given the equation sin(x + y)/sin(x - y)
Using the trigonometric identities.
sin(x + y) = sinxcosy + cosxsiny andsin(x - y) = sinxcosy - cosxsinySo, sin(x + y)/sin(x - y) = (sinxcosy + cosxsiny)/(sinxcosy + cosxsiny)
Dividing the rnumerator and denominator of ight hand side by sinx, we have
sin(x + y)/sin(x - y) = (sinxcosy + cosxsiny)/sinx/(sinxcosy + cosxsiny)/sinx
sin(x + y)/sin(x - y) = (sinxcosy/sinx + cosxsiny/sinx)/(sinxcosy/sinx + cosxsiny/sinx)
= (cosy + cotxsiny)/(cosy + cotxsiny) (since cosx/sinx = cotx)
Dividing the numerator and denominator of the right hand side by cosy, we have
= (cosy + cotxsiny)/cosy/(cosy + cotxsiny)/cosy
= (cosy/cosy + cotxsiny/cosy)/(cosy/cosy + cotxsiny/cosy)
= (1 + cotxstany)/(1 + cotxtany) [since siny/cosy = tany]
So, sin(x + y)/sin(x - y) = (1 + cotxstany)/(1 + cotxtany)
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