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2 years ago

How many 3/4's are in the whole number 21?

Mathematics

2 answers:

Nonamiya [84]2 years ago

7 0

Answer:

21 ÷ 3/4

= 21 × 4/3

= 7×4

= 28

Step-by-step explanation:

lora16 [44]2 years ago

3 0

Answer: 15.75

Step-by-step explanation:

3/4*21=15.75

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7 less than the product of a number n and 1/6 is no more than 98

VikaD [51]

Answer:

1/6, 1/6, 630

Step-by-step explanation:

1/6*n-7≤98

1/6*n≤98+7

1/6*n≤105

n≤105*6

n≤630

7 0

2 years ago

What is 3x=5 a positive solution, negetive solution, zero solution, or no solution

Margaret [11]

No solution because 3rimes something doesn't equal 5

3 0

2 years ago

The figure is cute into 12 equal pieces shade 3/4 of the figure

netineya [11]

Shade in 9 boxes

12÷4= 3 (This is 1 fourth)

3×3=9 (This is 3 fourths)

6 0

3 years ago

Write the sum in sigma notation. 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50

alisha [4.7K]

The sum in sigma notation for the sequence will be as follows:
From
<span>5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50
first term=5
common difference=5
number of terms=10
n=nth term
thus the sum will be:
(i=2 to 10)</span>∑(5(n-1)+5)

8 0

2 years ago

Read 2 more answers

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