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3 years ago

10+1/2(4x-8)^2 what is f(3)

1 answer:

5 0

$f(x) = 10 + \frac{1}{2} {(4x - 8)}^{2} \\ \\ f(3) = 10 + \frac{1}{2} {(4 \times 3 - 8)}^{2} \\ \\ f(3) = 10 + \frac{1}{2} {(12 - 8)}^{2} \\ \\ f(3) = 10 + \frac{1}{2} {(4)}^{2} \\ \\ f(3) = 10 + \frac{1}{2} \times 16 \\ \\ f(3) = 10 + 8 \\ \\ \boxed{ \red{ \large{f(3) = 18}}}$

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Find the ratio of:<br> (a) 5 km to 400 m<br> (b) 2 hours to 160 minutes

aleksandr82 [10.1K]

Answer:

Answer - a) 1:80

b) 1:80

Step-by-step explanation:

All the ratio should be written in lowest form. these are the answers for both the questions.

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2 years ago

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Two is subtracted from a number, and then the difference is multiplied by 5. The result is 30

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3 years ago

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Solve for s. 2/3 s + 5/6 s = 21 Enter the answer, as a whole number or as a fraction in simplest form, in the box.

dem82 [27]

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3 years ago

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The sides of a parallelogram are 40 and 70 feet long,and the smaller angle has a measure of 36.Find the length of the longer dia

Mumz [18]

Answer:

105 feet.

Step-by-step explanation:

The longer diagonal is opposite the longer angle in the parallelogram.

The longer angle = 180 - 36 = 144 degrees. (Reason: he same side angles are supplementary).

So, by the law of cosines:

x^ 2 = 40^2 + 70^2 - 2*40*70 cos 140 where x = the diagonal.

x^2 = 11030.50

x = 105.03 feet.

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3 years ago

A sphere of radius r is cut by a plane h units above the equator, where

Anika [276]

Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$

5 0

3 years ago