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3 years ago

Given h(t)=-2(t+5)^2+4, find h(-8) -334 -14 -45 -445

1 answer:

6 0

Substitute -8 for t in the expression, and now you have: h(-8) = -2[(-8) + 5]^2 + 4. Solve inside the parentheses; add -8 and 5.

h(-8) = -2[-3]^2 + 4, solve exponents next by squaring -3.

h(-8) = -2[9] + 4, multiply -2 and 9.

h(-8) = [-18] + 4, add -18 and 4.

h(-8) = -14.

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Consider the expression: (x + 3)•(y + 1)•(x + 2)

emmainna [20.7K]

Answer:

Step-by-step explanation:

First we multiply x through (y+1). This equals xy +x.

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3 years ago

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2 years ago

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Farmer bob's square plot ofland is slowly eroding away. Worried about the future of his farm. Farmer Bob measures the rate of er

kow [346]

Answer:

Option A.

Step-by-step explanation:

Area of a square is

$A=x^2$ .... (1)

where, x is side length.

The length of each side of his square plot is decreasing at the constant rate of 2 feet/year.

$\dfrac{dx}{dt}=2$

It is given that bob currently owns 250,000 square feet of land.

Fist find the length of each side.

$A=250000$

$x^2=250000$

Taking square root on both sides.

$x=500$

Differentiate with respect to t.

$\dfrac{dA}{dt}=2x\dfrac{dx}{dt}$

Substitute x=500 and $\frac{dx}{dt}=2$ in the above equation.

$\dfrac{dA}{dt}=2(500)(2)$

$\dfrac{dA}{dt}=2000$

Farmer bob is losing 2,000 square feet of land per year.

Therefore, the correct option is A.

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2 years ago

Consider the differential equation

Ainat [17]

Answer:

$W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )=e^x$

Step-by-step explanation:

Let $y=e^{\frac{x}{2}}$

Differentiate with respect to $x$

$y'=\frac{1}{2}e^{\frac{x}{2}}$

Differentiate with respect to $x$

$y''=\frac{1}{4}e^{\frac{x}{2}}$

Put values of $y,y',y''$ in $4y''-4y'+y=0$

$4y''-4y'+y=0\\4\left (\frac{1}{4}e^{\frac{x}{2}} \right )-4\left ( \frac{1}{2}e^{\frac{x}{2}}\right )+e^{\frac{x}{2}}\\=e^{\frac{x}{2}}-2e^{\frac{x}{2}}+e^{\frac{x}{2}}\\=2e^{\frac{x}{2}}-2e^{\frac{x}{2}}\\=0$

So, $y=e^{\frac{x}{2}}$ is the solution of the given equation.

Now, let $y=xe^{\frac{x}{2}}$

Differentiate with respect to $x$

$y'=e^{\frac{x}{2}}+\frac{x}{2}e^{\frac{x}{2}}=e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )$

Differentiate with respect to $x$

$y''=\frac{1}{2}e^{\frac{x}{2}}+\frac{1}{2}e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )=e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}}$

Put values of $y,y',y''$ in $4y''-4y'+y=0$

$4y''-4y'+y=0\\4\left (e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}} \right )-4\left ( e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )\right )+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-2e^{\frac{x}{2}}(2+x)+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-4e^{\frac{x}{2}}-2xe^{\frac{x}{2}}+xe^{\frac{x}{2}}\\=0$

To find: $W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )$

Solution:

Let $u=e^{\frac{x}{2}}\,,\,v=xe^{\frac{x}{2}}$

$W(u,v)=\left | \begin{matrix}u&v\\u'&v' \end{matrix} \right |\\=\left | \begin{matrix}e^{\frac{x}{2}}&xe^{\frac{x}{2}}\\\frac{1}{2}e^{\frac{x}{2}}&e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \end{matrix} \right |\\=e^{\frac{x}{2}}\left [ e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \right ]-\frac{1}{2}e^{\frac{x}{2}}xe^{\frac{x}{2}}\\=e^x\left ( 1+\frac{x}{2} \right )-\frac{1}{2}xe^x\\=e^x+\frac{1}{2}xe^x-\frac{1}{2}xe^x\\=e^x$

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2 years ago

2x - y = -2 x + y=-4

sertanlavr [38]

2x-y = - 2

2X - 0 = -2
2X = - 2
X = -2/2
X = -1

X + y = - 4

- 1

5 0

3 years ago