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anyanavicka [17]
2 years ago
11

PLEASE HELP ME! Select the correct answer. Consider function g. Which function could be the inverse of function g?

Mathematics
1 answer:
maria [59]2 years ago
8 0

Answer:

A)  k(x)

Step-by-step explanation:

An inverse function does the opposite to the function.

For example, if the function was "+2" then the inverse would be "-2", or if the function was "×2" the inverse would be "÷2".

An <u>inverse function</u> is a <u>reflection of the function in the line y = x</u>

Therefore (x, y) → (y, x)

(-8, 2) → (2, -8)

(-4, 3) → (3, -4)

(0, 4) → (4, 0)

(4, 5) → (5, 4)

(8, 6) → (6, 8)

Therefore, the function k(x) could be an inverse of function g(x).

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55 + 80 + 95 + 100 +115 +90 =535 actual enrollments. 75 + 80 +85 +90 +95 + 100 = 525 predicted enrollments. 535 - 525 = 10 residuals.
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. Sara teaches aerobics classes as a part-time job. The amount of money she earned is shown in the table below. Based on the inf
nikitadnepr [17]

Answer:

She should earn 225 dollars.

Step-by-step explanation:

I know this because 50 ÷ 4 = 12.5 and 12.5 × 18 = 225.

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2 years ago
Which figures have rotation symmetry?<br> Select each correct answer
goblinko [34]

Answer:

The both on the right are the correct answers

Step-by-step explanation:

This is because rotational symmetry is defined as the property a shape has when it looks the same after some rotation by partial turns.

8 0
3 years ago
1. The graph of quadratic function h is shown below. What are the zero(s) of h?
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6 0
2 years ago
Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators
Nata [24]

Answer:

The solution is y=-\frac{12}{12x-30x^2+1}.

Step-by-step explanation:

A first order differential equation y'=f(x,y) is called a separable equation if the function f(x,y) can be factored into the product of two functions of x and y:

f(x,y)=p(x)h(y)

where p(x) and h(y) are continuous functions.

We have the following differential equation

y'=(1-5x)y^2, \quad y(0)=-12

In the given case p(x)=1-5x and h(y)=y^2.

We divide the equation by h(y) and move dx to the right side:

\frac{1}{y^2}dy\:=(1-5x)dx

Next, integrate both sides:

\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C

Now, we solve for y

-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}

We use the initial condition y(0)=-12 to find the value of C.

-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}

Therefore,

y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}

4 0
3 years ago
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