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2 years ago

I have 9 questions please think wisely then comment your answers quick:

Mathematics

1 answer:

Viktor [21]2 years ago

4 0

Answer:

1.) 6 5/6+ 6 11/12= 13 3/4

2.) 4 2/3 + 3 3/4 = 8 5/12

3.) 1 1/2 + 2 3/10 = 3 4/5

4.) 4 1/6 + 5 4/7 = 9 31/42

5.) 4 1/6 + 5 4/7 = 9 31/42

6.) 3 8/9 + 1 5/12 = 5 11/36

7.) 3 1/4 + 2 1/6 = 5 5/12

8.) 4 9/16 + 1 5/8 = 6 3/16

9.) 2 1/8 + 1 3/4 = 3 7/8

Step-by-step explanation:

Idk, i just used symnolab for the most part

help a sistah, follow the instah: emventures666

plz give brainliest-

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Two-thirds of the difference between a number and four is six

hodyreva [135]

2/3 times x-4 = 6
thats the answer :)

7 0

2 years ago

The length of the base of a rectangular prism is 9

Tatiana [17]

Answer:

The volume of the prism is 320 15/16 inches³.

Step-by-step explanation:

Given:

The length of the base of a rectangular prism is 9 7/8.

The width of the prism is 8 1/8.

And the height is 4 inches.

Now, to find the volume of the prism.

(Length) l = 9 7/8 = 79/8 inches.

(Width) w = 8 1/8 = 65/8 inches.

(Height) h = 4 inches.

So, by putting the formula to get the volume:

Volume = w×h×l.

$Volume=\frac{65}{8}\times 4\times \frac{79}{8}$

$Volume=320\frac{15}{16}$

Volume = 320 15/16 inches³.

Therefore, the volume of the prism is 320 15/16 inches³.

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

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