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2 years ago

The width of a rectangle painting is 3 in. more than twice the height. A frame that is 2.5 in. wide goes around the painting.

Mathematics

1 answer:

Black_prince [1.1K]2 years ago

5 0

A.
$A=(2h+8)(h+5)$

b.
If h = 12
$A=(24+8)(12+5)=(32)(17)=544$

c.
If h = 15
$A=(30+8)(15+5)=(38)(20)=760$

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The mass of Earth is 5.97 >< 1024 kilograms. The mass of the Moon is 7.34 x 1022 kilograms. What is the combined mass of E

LuckyWell [14K]

Answer:

7.34x10^22 kg can also be written as 0.0734 * 10^24 kg

Since both numbers are * 10^24, you can just add 5.97 + 0.0734 = 6.043. Then you have to add the * 10^24 giving you 6.043 * 10^24 kg

The ratio of Earth / moon is 5.97x10^24 / 7.34x10^22 = 81. The mass of the Earth is 81 times the mass of the moon.

Step-by-step explanation: give me brainliest please :)

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3 years ago

What are the domain and range of the function

atroni [7]

Answer:

Domain (-∞,∞)

Range (0, ∞)

Step-by-step explanation:

The domain of a function is always the set of all real numbers therefore the domain is going to be (-∞,∞). The range is (0, ∞) because 0 is one of the outputs for the function

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3 years ago

Dolbear's law states the relationship between the rate at which Snowy Tree Crickets chirp and the air temperature of their envi

erastovalidia [21]

25.21 this is corect :D

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3 years ago

Simplify 1/3 + 1/5 - 1/7 divided by 1/x

fgiga [73]

41x
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3 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

2 years ago