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2 years ago

Find value of x round answer to nearest tenth

Mathematics

1 answer:

Aloiza [94]2 years ago

4 0

Answer:

Step-by-step explanation:

19^2 = y^2+ 17^2

y = 8.5

by figure

x = 31 - 2*y

x = 14

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2(9x - 1) = 99-7(3 - 4x)<br> help please

Bogdan [553]

Answer:

x = -8

Step-by-step explanation:

Distribute on both sides

combine constants

combine variables

then divide

6 0

3 years ago

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What is the solution to the system of equations shown in the graph?

Katen [24]

The solution is where both lines intersect. First find the x-coordinate which we will find by going to the left two units. That means x = -2. Now going up one unit, which gives us y = 1

(-2, 1)

8 0

3 years ago

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noname [10]

Answer:

287.1 inches of the canvas.

Step-by-step explanation:

To solve this, we need to first figure out the total area of the canvas. To do that, multiply width by height.

29*33=957

Now set up your equation for solving for the area of the canvas that the rose covers.

x/957=30/100

We did it where: x is the area of the rose covers, 957 is the amount of inches that the canvas takes up, and the right side of the equation is the percent.

Now cross multiply.

100x=28,710

Now divide both sides by 100.

x=287.1

The red rose covers 287.1 inches of the canvas.

6 0

3 years ago

Need some help...........

Anarel [89]

Answer:

The answer is to the question is 13

7 0

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