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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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In order for the parallelogram to be a<br> rhombus, x = [?].<br> (5x + 25)<br> (12x + 11)

LiRa [457]

A parallelogram is also a rhombus if the diagonal is a bisector of an angle enclosed by the two adjacent sides of a parallelogram.

In our case it means,

$5x+25=12x+11$

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3 years ago

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3 years ago

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3 0

3 years ago

1 Point

I am Lyosha [343]

Option C

The ratio for the volumes of two similar cylinders is 8 : 27

<h3><u>Solution:</u></h3>

Let there are two cylinder of heights "h" and "H"

Also radius to be "r" and "R"

$\text { Volume of a cylinder }=\pi r^{2} h$

Where π = 3.14 , r is the radius and h is the height

Now the ratio of their heights and radii is 2:3 .i.e

$\frac{\mathrm{r}}{R}=\frac{\mathrm{h}}{H}=\frac{2}{3}$

<em><u>Ratio for the volumes of two cylinders</u></em>

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{\pi r^{2} h}{\pi R^{2} H}$

Cancelling the common terms, we get

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{\mathrm{r}}{R}\right)^{2} \times\left(\frac{\mathrm{h}}{\mathrm{H}}\right)$

Substituting we get,

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{2}{3}\right)^{2} \times\left(\frac{2}{3}\right)$

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{2 \times 2 \times 2}{3 \times 3 \times 3}$

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{8}{27}$

Hence, the ratio of volume of two cylinders is 8 : 27

7 0

3 years ago

How to solve this problem

rodikova [14]

$\bf \textit{circumference of a circle}\\\\
C=2\pi r\quad 
\begin{cases}
r=radius\\
-----\\
r=10
\end{cases}\implies C=2\pi 10\implies \boxed{C=20\pi }\\\\
-------------------------------\\\\$

$\bf \textit{let's converte 1km to inches}
\\\\\\
km\cdot \cfrac{1000m}{km}\cdot \cfrac{100cm}{m}\cdot \cfrac{in}{2.54cm}\implies \cfrac{1000\cdot 100\cdot in}{2.54}\approx 39370.079\ in\\\\
-------------------------------\\\\
\textit{how many times does }20\pi \textit{ go into }39370.079?\implies \boxed{\cfrac{39370.079}{20\pi }}$

3 0

3 years ago