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3 years ago

30 POINTSS!!

1 answer:

4 0

F(x) = x^3 ... parent cubic function
g(x) = f(x-4) ... shifting f(x) 4 units to the right
g(x) = (x-4)^3
h(x) = g(x) + 6 ... shifting g(x) 6 units up
h(x) = (x-4)^3 + 6

The turning point of f(x) is at (0,0)
Applying the shifting of 4 units to the right and 6 units up, we get to (4,6) as the new turning point for h(x). This also the turning point for T(x) as well.

What does it mean? Well that part isn't clear to me because its not clear what x and T(x) represent. I'm assuming T(x) is temperature though I don't know what x represents. It's possible that x represents time though again I'm not sure about this part.

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Solve the following system: y = x 3 4x y = 18 (6, 3) (3, 6) (â’3, 6) (3, â’6).

Nastasia [14]

The solution of the given equation is (3,6). The correct option is B.(3,6)

<h3>Given equations,</h3>

$y=x+3.......(1)\\4x+y=18.....(2)$

<h3>How to solve the equations?</h3>

putting the value of y from equation (1) in equation (2) , we get

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So, the correct option is B.

For more details about the system of equations, follow the link:

brainly.com/question/12895249

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2 years ago

the half-life of strontium-90 is approximately 29 years. how much of a 500 g sample of strontium-90 will remain after 58 years

Yuliya22 [10]

Answer: 125 g

<u>Step-by-step explanation:</u>

$A = P_o\cdot e^{kt}\\\\\text{First, use the given information to find k:}\\\\\bullet A=\dfrac{1}{2}P_o\\\\\bullet k = unknown\\\\\bullet t=29\text{ years}\\\\\dfrac{1}{2}P_o=P_o\cdot e^{k(29)}\\\\\\\dfrac{1}{2}=e^{k(29)}\qquad divided\ both\ sides\ by\ P_o\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=ln\bigg(e^{k(29)}\bigg)\qquad applied\ ln\ to\ both\ sides\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=29k\qquad simplified-ln\ and\ e\ cancel\ out\\\\\\\dfrac{ln\bigg(\dfrac{1}{2}\bigg)}{29}=k\qquad divided\ 29\ from\ both\ sides\\\\\\-0.0239=k$

$\text{Now, use the following in the equation to solve for A:}\\\\\bullet A=unknown\\\bullet P_o=500\\\bullet k=-0.0239\\\bullet t=58\text{ years}\\\\A=500\cdot e^{(-0.0239)(58)}\\\\.\quad=500\cdot e^{-1.386}\\\\.\quad=125$

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