Answer: y= -2x -2
Step-by-step explanation:
1) Create an equation going through (1,-4)
Replace 5 with b otherwise, we can't find an equation that goes through it.
y = -2x + b
2) Substitute y and x with the point. Solve.
-4 = -2(1) + b
-4 = -2 + b
-2 = b
3) Reput in the equation
y= -2x -2
Note: It is parallel because they have the same slope!
The cost of creating a software program is $5000. Every extra feature added to the software costs $100. Thetotal charge of the software with x extra features is given by the function f(x) = 100x + 5000. How will the graphof this function change if the basic cost is raised to $5200 and the cost of each extra feature is increased to$120?
The original function is:
[tex]f(x)=100x+5000[/tex]Which means that it has a slope of 100, and a starting point of 5000.
As we can see on the image above, the first point is (0,5000), which is the starting point of the function. The second point is (2, 5200), which means that for a 2 unit increase on the features, we got 200 increase on the price, leading to a slope of 100.
If we change the basic cost to 5200, the starting point of the function will increase by 200 units, if we change the slope to 120, then for every feature we add, the cost will increase faster. In summary, the cost of the program will be higher to start, and it will also increase faster. This is illustrated in the image below:
[tex]g(x)=120x+5200[/tex]In green we have the new function, and in red we have the old one. We can see that the new one increases faster, and is always above the old one.
help help help pleaseeeee!!!!!!
a) The linear equation that models the price-sales relationship for toy is C(x) = -500x +5500
b) The forecast calls for 2250 sales at a $6.50 pricing.
Define slope.The ratio of the increase in elevation between two points to the run in elevation between those same two points is referred to as the slope.
A line's equation is represented by:
y = mx +b
, where
The slope, or m, represents the rate of change.
The value of y at x = 0 is represented by the y-intercept or b.
Item a:
In this issue:
Two points are (6, 2500) and (8, 1500).
The slope is calculated by dividing the change in y by the change in x, so:
m = [tex]\frac{1500-2500}{8-6}[/tex]
m = [tex]\frac{-1000}{2}[/tex]
m = -500
Thus,
y = -500x +b
Point (6,2500) indicates that, which we utilize to find b, is true when.
y = -500x +b
2500 = -500(6) + b
b = 5500
Thus
y = -500x + 5500
Item b:
When x = 6.5, sales are y, so:
y = -500(6.5) + 5500
y = 2250
The forecast calls for 2250 sales at a $6.50 pricing.
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when Ibuprofen is given for fever to children 6 months of age up to 2 years, the usual dose is 5 milligrams (m) per kilogram (kg)of body weight when the fever is under 102.5 degrees Fahrenheit how much medicine would be usual dose for a 18 month old weighing 20 poundsRound your answer to the nearest milligram
dose: 5 milligrams (m) per kilogram (kg)
1 pound= 0.5kg
20 pounds= 10kg
[tex]5\text{ m/kg }\cdot10\operatorname{kg}=50m[/tex]the usual dose for a 18 month old weighing 20 pounds is 50 miligrams
Graph the exponential function.G(x)=(1/3)^xPlot five point on the graph of the function.
Graph the function:
[tex]G(x)=(\frac{1}{3})^x[/tex]We'll use the following values of x: {-2, -1, 0, 1, 2}.
Substituting:
[tex]G(-2)=(\frac{1}{3})^{-2}=3^2=9[/tex][tex]G(-1)=(\frac{1}{3})^{-1}=3^1=3[/tex][tex]G(0)=(\frac{1}{3})^0=1[/tex][tex]G(1)=(\frac{1}{3})^1=0.333[/tex][tex]G(2)=(\frac{1}{3})^2=0.111[/tex]The graph of the function is shown below:
write an expression in simplest form for the perimeter of a right triangle with leg lengths of 12a^4 and 16a^4an expression is____a^4
Given the lengths of the right triangle:
[tex]\begin{gathered} 12a^4^{} \\ \text{and } \\ 16a^4 \end{gathered}[/tex]To find the perimeter, use the formula:
[tex]P=a+b+\sqrt[]{a^2+b^2}[/tex]Thus, we have:
[tex]P=12a^4+16a^4+\sqrt[]{(12a^4)^2+(16a^4)^2}[/tex][tex]\begin{gathered} P=28a^4+\sqrt[]{144a^4+256a^4} \\ P=28a^4+\sqrt[]{400a^4} \\ P=28a^4+20a^4 \\ P=48a^4 \end{gathered}[/tex]Therefore, the expression is:
[tex]48a^4[/tex]Pls help this is geometry and its a really important test
Answer:#4:B m of angle 2=35° m of angle 3=55°
Step-by-step explanation:if line C is 180° degrees because a flat line in 180° then you see that adding the angle measure of 4 which is 35° +the right angle which is 90 degree=125° So the 180°-125°=55°=m of angle 3 and to find angle measure of 2 angle 1 is a right angle and the inside of a triangle opens to 180 degrees so you grab measure of angle 1=90° and measure of angle 3=55° add them 90°+55°=145° so 180°-145°=35°
The sum of the even numbers between 77 and 535 is how much less than the sum of the odd numbers between78 and 536?
The sum of the even numbers between 77 and 535 is:
[tex]78+80+82+\ldots+530+532+534=70074[/tex]Now, the sum of the odd numbers between 78 and 536 is:
[tex]77+79+81+\cdots+531+533+535=70380[/tex]Thus, the sum of the even numbers between 77 and 535 is 306 less than the sum of the odd numbers between 78 and 536 because
[tex]70380-70074=306[/tex]Hello! I just want to confirm my answer is correct and if there’s anything I should add? Thanks for your help!
Given:
given functions are
[tex]f(x)=x^4-4x^3-2x^2-12x+9,g(x)=\sqrt{x^2-2x-3},h(x)=\frac{-x^2+1}{x^2-2x-3}[/tex]Find:
(A) we have to compare the Domain and range of the function f(x) and g(x).
(B) We have to find the relationship between the break of h(x) and zeros of f(x).
Explanation:
The domain and Range of f(x) is
[tex]\begin{gathered} Domain(f)=(-\infty,\infty) \\ Range(f)=[0,\infty) \end{gathered}[/tex]Domain and Range of g(x) is
[tex]\begin{gathered} Domain(g)=(-\infty,-1]\cup[3,\infty) \\ Range(g)=[0,\infty) \end{gathered}[/tex]Domain of h(x) is
[tex]Domain(h)=(-\infty.-1)\cup(-1,3)\cup(3,\infty)[/tex]Now zeros of the function f(x) are
[tex]\begin{gathered} x^4-4x^3-2x^2+12x+9=0 \\ (x+1)^2(x-3)^2=0 \\ x=-1,-1,3,3 \end{gathered}[/tex]Therefore, zeors of the function f(x) are -1,-1,3,3.
Now,
(A)The difference between Domain of f(x) and g(x) is of the interval (-1,3). The domain of f(x) is all Real Numbers and Domain of g(x) is all the real number except the interval (-1,3).
The Range of both f(x) and g(x) is same.
(B) The breaks in the Domain of h(x) are equal to the zeros -1, 3 of f(x).
please help me out quickly
An identity matrix is a square matrix with 0s everywhere else and 1s on the major diagonal.
1. The given statement exists true.
2. The given statement exists false.
What is an inverse matrix transform?A rigid body transformation is the inverse matrix. It not only conforms to the original matrix's shape, but if you translate and rotate an object, you may return it to its original location by undoing the translations and rotations.
In general, AB = BA, even if A and B exist as both squares. If AB = BA, then we express that A and B commute. For a general matrix A, we cannot express that AB = AC yields B = C. (However, if we comprehend that A exists invertible, then we can multiply both sides of the equation AB = AC to the left by A-1 and get B = C.)
2. The given statement exists false.
An identity matrix is a square matrix with 0s everywhere else and 1s on the major diagonal.
Identity matrices of orders 1, 2, and 3 are those in which all the members of a square matrix are zero and none of the diagonal elements are zero. Keep in mind that when k = 1, a scalar matrix is an identity matrix. But it is obvious that any identity matrix is a scalar matrix.
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The first statement is true but the second given statement is false .
Inverse of a matrix: What does that mean?The idea of the inverse of a matrix is a multidimensional generalization of the idea of the reciprocal of a number. Just as the product of a number and its inverse is equal to 1, that of a square matrix and its inverse is equal to the identity matrix.
Even though A and B are squares, AB Equals BA in general. We can say that A and B commute if AB = BA. We cannot state that AB = AC produces B = C for a general matrix A. (However, if we are aware that A is invertible, we can multiply both sides of the expression AB = AC to the left by A-1 to obtain B = C.)
2. The aforementioned claim is untrue.
A square matrix with 1s on the principal diagonal and 0s everywhere else is known as an identity matrix.
When all of a square matrix's members are zero and none of the diagonal elements are zero, the matrix is said to be an identity matrix of order 1, 2, or 3. Do not forget that k = 1 .
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Kamal drew a scale drawing of a theater. He used the scale 6 inches = 10 feet. What scale factor does the drawing use?Simplify your answer and write it as a ratio, using a colon.
Given:
Kamal drew a scale drawing of a theater.
He used the scale 6 inches = 10 feet.
The scale factor will take the form:
[tex]inches\colon feet=x\colon y[/tex]Where (x) is the number of inches that corresponding to the number of feet (y)
So, x = 6, y = 10
So, the scale factor =
[tex]inches\colon feet=6\colon10[/tex]Simplifying the ratio, so the answer will be:
[tex]inches\colon feet=3\colon5[/tex]Anyone know how to work this?
The income from the sales of potatoes in 2021 is equal to €510.
How to calculate the income from potatoes?From the information provided, we can logically deduce that the total income (revenue) that was generated from the sales of the four (4) vegetables in 2021 is equal to 2040 Euros (€).
Note: Each interval of the bar chart on the y-coordinate represents €40.
Assuming the pie chart was drawn to scale, we can reasonably infer and logically deduce that each of its sector represent 25%. This ultimately implies that, the income for the sales of potatoes in 2021 is 0.25 of the total income (revenue) that was generated from the sales of the four (4) vegetables in 2021.
Mathematically, the income for the sales of potatoes in 2021 can be calculated as follows:
Income from potatoes = 25/100 × 2040
Income from potatoes = 0.25 × 2040
Income from potatoes = €510.
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Which of the following is the graph of y = |3+x|?
Which characteristic of the line that passes through the points (6,10) and (12,2)..
SOLUTION:
Let us find the slope of the line:
[tex]m\text{ = }\frac{y_2-y_1}{x_2-x_1}[/tex]Where x1 = 6, x2 = 12, y1 = 10 and y2 = 2
[tex]\begin{gathered} m\text{ = }\frac{2-10}{12-6} \\ m\text{ = }\frac{-8}{6}\text{ = -}\frac{4}{3} \\ \\ \end{gathered}[/tex]The slope of the given line is -4/3
It takes 6 slices of bread, 9 oz of cheese, and 2 oz of butter tomake three grilled-cheese sandwiches. What is the cost per sandwichif bread (18 slices) costs $1.90, 1 lb of cheese costs $2.49, and1/2 lb of butter costs $1.29?a. $.78b. $4.87c. $2.12d. $1.62
Step 1: Write the equivalence of each unit
[tex]\begin{gathered} 1lb\Rightarrow16\text{ oz} \\ 1\text{ lb of cheese }\Rightarrow\text{ 16 oz of che}ese \\ \frac{1}{2}\text{ lb of butter }\Rightarrow\text{ 8 oz of butter} \end{gathered}[/tex]Step 2: Calculate the cost of producing 3 grilled-cheese sandwiches
[tex]\begin{gathered} 18\text{ slices of bread cost \$1.90} \\ 1\text{ slice will cost x} \\ \Rightarrow x=\frac{1.90}{18} \\ 6\text{ slices of bread will cost=}\frac{1.90}{18}\times6=\text{ \$0.63} \end{gathered}[/tex][tex]\begin{gathered} 1\text{ lb of cheese cost \$2.49 } \\ \text{ Since 1 lb is equal to 16 oz} \\ 16\text{ oz of cheese cost \$2.49} \\ 1\text{ oz of cheese will cost \$x} \\ \Rightarrow\text{ \$x=}\frac{2.49}{16}\text{ } \\ 9\text{ oz will cost }\Rightarrow\text{ }\frac{2.49}{16}\times9=\text{ \$1.40} \end{gathered}[/tex][tex]\begin{gathered} \frac{1}{2}\text{ lb of butter cost \$1.29} \\ \text{ This implies } \\ 8\text{ oz of butter will cost \$1.29} \\ 1\text{ oz of butter will cost \$x} \\ \Rightarrow\text{ \$x =}\frac{1.29}{8} \\ 2\text{ oz will then cost =}\frac{1.29}{8}\times2=\text{ \$0.32} \\ \end{gathered}[/tex]From the above calculations we can calculate the cost of producing three grilled-cheese sandwiches as
[tex]0.63+1.40+0.32=\text{ \$2.35}[/tex]Thus, the cost per sandwich is given as
[tex]\frac{2.35}{3}=\text{ \$0.78}[/tex]Hence, the cost per sandwich is $0.78
Option A is the right answer
Please help I don't understand how to do this. I am stuck
For this exercise, you will demonstrate the tests to show that y=x^19-8x^7+7x^4 is neither an even function nor an odd function.
A) When you apply the test for evenness and simplify the resulting equation, you get
y=
B) When you apply the test for oddness and simplify the resulting equation, you get
y=
The given equation is proved that y=x¹⁹-8x⁷+7x⁴ is neither an even nor an odd function.
Given the equation is y=x¹⁹-8x⁷+7x⁴
Replace x and -x and check to see if the resultant equation matches the
original equation to see whether the function is even
Write the original equation into a function first by swapping y for f(x).
f(x) = x¹⁹-8x⁷+7x⁴
now, replace x with -x
f(-x) = (-x)¹⁹-8(-x)⁷+7(-x)⁴
f(-x) = -x¹⁹+8x⁷+7x⁴
now check whether f(x) = f(-x)
since f(x)≠f(-x), the function is not even.
If you want to determine whether a function is odd, you should check to see if f(-x) = -f (x). The function is odd if the two equations are the same.
-f(x) = -(x¹⁹-8x⁷+7x⁴)
-f(x) = -x¹⁹+8x⁷-7x⁴
now check whether f(-x) = -f(x)
since f(-x)≠-f(x), the function is not odd.
hence the equation is neither an even nor an odd function.
Therefore, the equation is proved that it is neither even nor odd.
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Solve the following for 'x'x+1+3 2= xO A. -1O B. 3O c. 3OD. 1
Step 1: "braking" fractions
[tex]\begin{gathered} \frac{x}{3}+\frac{x+1}{2}=x \\ \frac{x+1}{2}=\frac{x}{2}+\frac{1}{2} \end{gathered}[/tex]We replace the second in the original equation:
[tex]\frac{x}{3}+\frac{x}{2}+\frac{1}{2}=x[/tex]Step 2: rearraging the equation (the terms with x on one side, numbers on the other)
[tex]\begin{gathered} \frac{x}{3}+\frac{x}{2}+\frac{1}{2}=x \\ \frac{x}{3}+\frac{x}{2}=x-\frac{1}{2} \\ \frac{x}{3}+\frac{x}{2}-x=-\frac{1}{2} \end{gathered}[/tex]Step 3: adding fractions
Since
[tex]\begin{gathered} \frac{1}{3}+\frac{1}{2}-1=\frac{1}{3}+\frac{1}{2}-\frac{1}{1} \\ =(\frac{1}{3}+\frac{1}{2})-\frac{1}{1} \\ \end{gathered}[/tex]We know that
[tex]\begin{gathered} (\frac{1}{3}+\frac{1}{2})=\frac{1\cdot2+1\cdot3}{3\cdot2} \\ =\frac{2+3}{6} \\ =\frac{5}{6} \end{gathered}[/tex]Replacing it:
[tex]\begin{gathered} (\frac{1}{3}+\frac{1}{2})-\frac{1}{1}=\frac{5}{6}-\frac{1}{1} \\ =\frac{5\cdot1-6\cdot1}{6\cdot1} \\ =\frac{5-6}{6} \\ =-\frac{1}{6} \end{gathered}[/tex]Then
[tex]\begin{gathered} \frac{x}{3}+\frac{x}{2}-x=-\frac{1}{2} \\ -\frac{1}{6}x=-\frac{1}{2} \end{gathered}[/tex]Step 4: finding x
[tex]\begin{gathered} -\frac{1}{6}x=-\frac{1}{2} \\ \frac{1}{6}x=\frac{1}{2} \\ 6\cdot\frac{1}{6}x=6\cdot\frac{1}{2} \\ x=3 \end{gathered}[/tex]Answer: C.x=3For each diagram, write and solve an inequality for x.
7.
2x - 5
X
What is the answer
For the given diagram the inequality for x is x ≥ 6.
What is inequality?If two values differ, it can be determined whether one is larger, smaller, or simply not equal to the other.
A and B claim that they are not equal.
A must be less than B if A = B.
An is greater than b when a > b. These two constitute strict inequality.
A must be less than or equal to b if a and b are divided.
An is greater than b if a and b are equal.
We have given a right-angled triangles with hypotenuse 2x - 5 and base x
Using the Pythagoras theorem
c² = a² + b²
We know hypotenuse is always greater than base with that in mind lets assume x to be 1
then 2x - 5
= 2(1)-5
= -3 which is smaller than 1
It means x at least need to be equal to 6 or greater than 6 to Pythagoras theorem to be possible
When x = 6
= 2(6)-5
= 7 hypotenuse
Gives the inequality
x ≥ 6
Thus, For the given diagram the inequality for x is x ≥ 6.
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Range for function y=-3x+12
The range of y=-3x+12 is (-∞,∞) or (y | y ∈ R)
What is Domain and Range ?
The components of a function are its domain and range. The domain of a function is the set of all possible input values, whereas the range of a function is its potential output. Range, Domain, and Function
Since no domain is specified to determine the precise range, As a result, the range of the supplied equation, y=-3x+12, only includes real integers.
Interval Notation : (−∞,∞)
The range is the set of all valid y values. You can also use the graph to find the range. The range is ,
Interval range : (−∞,∞)
Hence, The Range is (-∞,∞) or (y | y ∈ R)
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I am not understanding this question can you help me solve this? Please I desperately need your help
Given:
[tex]\[g\left(x\right)=\left\{\begin{matrix}\sqrt[3]{x+5} \\ -{{x}^2}+11\end{matrix}\text{ }\begin{matrix}x\le-4 \\ x>-4\end{matrix}\right.\][/tex]To find:
The value of:
[tex]g(-4)[/tex]Explanation:
For x = -4, we can consider the below function,
[tex]g(x)=\sqrt[3]{x+5}[/tex]Substituting the value x = -4,
[tex]\begin{gathered} g(-4)=\sqrt[3]{-4+5} \\ =\sqrt[3]{1} \\ =1 \end{gathered}[/tex]Final answer:
Hence, the required solution is:
[tex](A)\text{ 1}[/tex]if you do this I will name my future kid after you
We have to determine wheter the line segments MN and RS are parallel, perpendicular or neither. To do this we will find the slope of each of this segments; the slope is given by
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]For the line segment MN we have M(-2,2) and N(1,-3), then
[tex]\begin{gathered} m_1=\frac{-3-2}{1-(-2)} \\ =\frac{-5}{3} \end{gathered}[/tex]For the line segment RS we have R(-2,1) and (3,4), then
[tex]\begin{gathered} m_2=\frac{4-1}{3-(-2)} \\ =\frac{3}{5} \end{gathered}[/tex]Now that we have the slopes of each lines segments we have to remember two theorems.
T1. Two lines are parallel if and only if
[tex]m_1=m_2[/tex]T2. Two lines are perpendicular if and only if
[tex]m_1m_2=-1[/tex]Once we know this theorems we can answer the question.
First, we notice that the slopes of this segments are not equal so we can conlcude that they are not parallel.
Let's see if they are perpendicular, to do this we multiply the slopes
[tex]\begin{gathered} m_1m_2=(-\frac{5}{3})(\frac{3}{5}) \\ =-\frac{15}{15} \\ =-1 \end{gathered}[/tex]Since the result of their multiplication is -1 we conclude that this lines segments are perpendicular.
how to solve a given fraction by multiplying the denominators, not by factoring
In the picture there is a problem which is incomplete. There are two terms with the numerator and denominator. The fraction is [tex]\frac{a-b}{2a^{2}-ab-3b^{2} } -\frac{a+b}{2a^{2}-5ab+3b^{2}}=0[/tex]
Given that,
In the picture there is a problem which is incomplete.
We have to complete the fractions by solving.
There are two terms with the numerator and denominator.
There are only variable not numbers.
We have,
=[tex]\frac{a-b}{2a^{2}-ab-3b^{2} } -\frac{a+b}{2a^{2}-5ab+3b^{2}}[/tex]
We have to take an LCM.
[tex]\frac{(a-b)(2a^{2}-ab-3b^{2})- (a+b)(2a^{2}-5ab+3b^{2})}{(2a^{2}-ab-3b^{2})(2a^{2}-5ab+3b^{2})}[/tex]
Now,
Just take the numerator term and solve it
[tex](a-b)(2a^{2}-ab-3b^{2})- (a+b)(2a^{2}-5ab+3b^{2})[/tex]
Separate each term with multiplication
a(2a²-ab-3b²)-b(2a²-ab-3b²)-a(2a²-5ab+3b²)-b(2a²-5ab+3b²)
Multiply the terms
2a³-a²b-3ab²-2a²b+ab²+3b³-2a³+5a²b-3ab²-2a²b+5ab²-3b³
Arrange the terms to get calculation easy
2a³-2a³-a²b-2a²b+5a²b-2a²b-3ab²+ab²-3ab²+5ab²+3b³-3b³
Subtract the terms
2a³-2a³=0
-a²b-2a²b+5a²b-2a²b=-5a²b+5a²b=0
-3ab²+ab²-3ab²+5ab²=-6ab²+6ab²=0
3b³-3b³=0
Now, We get the numerator as 0.
So, If numerator is 0 then the whole term is 0.
Because, any term divides by 0 is 0.
Therefore, The fraction is [tex]\frac{a-b}{2a^{2}-ab-3b^{2} } -\frac{a+b}{2a^{2}-5ab+3b^{2}}=0[/tex]
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mrs. Baker phone 25 shells on the beach she brought 2/5 of the shells to our classroom how many shells did she bring to her classroom
Given that Mrs. Baker phone 25 shells on the beach, and the brought 2/5 of the shells to class.
[tex]\begin{gathered} \text{total T = 25} \\ \text{fraction brought to class }f(C)=\frac{2}{5} \end{gathered}[/tex]The number of shells she bring to class will be the product of the fraction and the total;
[tex]\begin{gathered} n(C)=f(C)\times T \\ n(C)=\frac{2}{5}\times25 \\ n(C)=10 \end{gathered}[/tex]Therefore, the number of shells she bring to class is;
[tex]10[/tex]27. The figure is formed from rectangles. Find the total area. The diagram is not to scale.3 ft5 ft3 ft8 ftA. 26 ft?B. 40 ftC. 73 ft?D. 34 ft?
To find the total area, we can divide the total figure into two figures as follows:
Then, we have that the total area is the sum of both areas of the figures:
[tex]A_{\text{total}}=5ft\cdot5ft+3ft\cdot3ft=25ft^2+9ft^2=34ft^2[/tex]Therefore, the total area is equal to 34 sq. feet (option D).
The volume of this triangular prism is 20,580 cubic millimeters. What is the value of p?
We have that the formula of the volume of the triangular prism is:
[tex]V=A\cdot h[/tex]Where A is the area of the base and h is the height. Then we can write the volume like this:
[tex]\begin{gathered} A=\frac{p\cdot b}{2} \\ \Rightarrow V=(\frac{p\cdot b}{2})\cdot h=\frac{p\cdot b\cdot h}{2} \end{gathered}[/tex]now, if b=20, h=42 and V=20580, then we substitute and solve for p:
[tex]\begin{gathered} V=\frac{p\cdot b\cdot h}{2} \\ \Rightarrow20580=\frac{p\cdot20\cdot42}{2}=420\cdot p \\ \Rightarrow p=\frac{20580}{420}=49 \\ p=49 \end{gathered}[/tex]therefore, the value of p is 49 milimeters
formulas need help will send picutre
In the formula
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_2}[/tex]putting in the given values of R_1, R_2, and R_3 gives
[tex]\frac{1}{R}=\frac{1}{10.0}+\frac{1}{14.0}+\frac{1}{35.0}[/tex]We add the fractions on the right-hand- side to get:
[tex]\frac{1}{R}=\frac{1\cdot7}{10.0\cdot7}+\frac{1\cdot5}{14.0\cdot5}+\frac{1\cdot2}{35.0\cdot2}[/tex][tex]\frac{1}{R}=\frac{7}{70}+\frac{5}{70}+\frac{2}{70}[/tex][tex]\frac{1}{R}=\frac{14}{70}=\frac{1}{5}[/tex]Taking the reciprocal both sides gives
[tex]R=5.0[/tex]Which is our answer!
simplify both questions and give full explanation please
Answer:
First question: 1/x^-7
Second question:
x^7
Step-by-step explanation:
To divide these problems (same variable base and exponents) subtract the exponents.
To get rid of a negative on the exponent, "push" the term across the fraction bar. Passing over the fraction bar changes the sign of the exponent. There are math reasons for this, its not random. But thats how it works. (Has to do with an exponent of -1 which will give you the reciprocal of your base).
Also, to multiply terms with the same base, add the exponents.
see image.
B / 7 equals 3 what is b
Answer:
B = 21
Step-by-step explanation:
B/7 =3
cross multiply
B = 21
Need Help on this problem
Answer:
Marginal Average Cost Function C'(x) = 5.7
Step-by-step explanation:
The Marginal Average Cost is just the first differential of the Cost Function
[tex]\dfrac{d}{dx} C(x) = \dfrac{d}{dx}(161) + \dfrac{d}{dx}(5.7x)\\\\\\The first differential of a constant is 0 and the first differential of ax is a \\\\So ,\\\\\dfrac{d}{dx} C(x) = 0 + 5.7 = 5.7[/tex]
This means that as x increases by 1 unit, cost increases by 5.7 units. x presumably refers to the number of units produced
The Marginal Revenue can be found the same way
[tex]\dfrac{d}{dx} R(x) = \dfrac{d}{dx}6x - \dfrac{d}{dx}(0.08x^2)\\\\\dfrac{d}{dx}6x = 6\\\\\dfrac{d}{dx}(0.08x^2) = (2)(0.08)x[/tex]
[tex]= 0.16x[/tex] since [tex]\dfrac{d}{dx} (x^2) = 2x[/tex]
Find the distance
d(P1.P₂) between
the given points P1
and P2.
P1 = (0,0)
P2 = (6,5)
Simplify your answer, use radicals as needed.
The distance between the two points P1 and P2 is d = √61.
What is the distance?Distance is a measurement of how far apart two objects or points are, either numerically or occasionally qualitatively. The distance can refer to a physical length in physics or to an estimate based on other factors in common usage. Length is measured in distance. For instance, the length of a road is its distance. The most popular units of measurement for distance in the metric system are millimeters, centimeters, meters, and kilometers.So, the distance between points P1 and P2:
Where, P1 = (0,0) and P2 = (6,5).The distance formula: d = √(x₂-x₁)²+(y₂-y₁)²Now, substitute the values in the formula as follows:
d = √(x₂-x₁)²+(y₂-y₁)²d = √(6-0)²+(5-0)²d = √36+25d = √61Therefore, the distance between the two points P1 and P2 is d = √61.
Know more about distance here:
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Find the distance between (5, -7) and (-2,-4)
Answer: 7.62
Step-by-step explanation:
