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4 years ago

15

Find the length of the rectangle whose area is 99 square inches and the width is 9 inches.

1 answer:

Oliga [24]4 years ago

7 0

Answer:

the answer is 11

Step-by-step explanation:

??×9=99

9÷99=11

11×9=99

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Can someone help me with this maths question

vazorg [7]

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3 years ago

The price of a shoes as reduced from 60 to 45. by what percentage was the price of the shoes reduced?

sweet-ann [11.9K]

45 is 75% of 60 but if you’re looking for the difference subtract 75 from 100, so 25%

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3 years ago

Gretchen has a set of blocks of heights 1, 2, and 4-centimeters. Imagine stacking the blocks one on top of the other to make a t

notsponge [240]

Answer:

$t_{n}$ = $t_{n-1}$ + $t_{n-2}$ + $t_{n-4}$

Step-by-step explanation:

$t_{n}$ =multiple ways to climb a tower

When n = 1,

tower= 1 cm

$t_{1}$ = 1

When n = 2,

tower =2 cm

$t_{2}$ = 2

When n = 3,

tower = 3 cm

it can be build if we use three 1 cm blocks

$t_{3}$ = 3

When n = 4,

tower= 4 cm

it can be build if we use four 1 cm blocks

$t_{4}$ = 6

When n > 5

tower height > 4 cm

so we can use 1 cm, 2 cm and 4 cm blocks

so in that case if our last move is 1 cm block then $t_{n-1}$ will be

n —1 cm

if our last move is 2 cm block then $t_{n-2}$ will be

n —2 cm

if our last move is 4 cm block then $t_{n-4}$ will be

n —4 cm

$t_{n}$ = $t_{n-1}$ + $t_{n-2}$ + $t_{n-4}$

4 0

4 years ago

Read 2 more answers

prohojiy [21]

The answer would be B, due to the fact that the shaded area is above the line.

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3 years ago

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo

oksano4ka [1.4K]

Answer:

Required center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

Step-by-step explanation:

Given semcircles are,

$y=\sqrt{1-x^2}, y=\sqrt{16-x^2}$ whose radious are 1 and 4 respectively.

To find center of mass, $(\bar{x},\bar{y})$ , let density at any point is $\rho$ and distance from the origin is r be such that,

$\rho=\frac{k}{r}$ where k is a constant.

Mass of the lamina=m= $\int\int_{D}\rho dA$ where A is the total region and D is curves.

then,

$m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k$

Now, x-coordinate of center of mass is $\bar{y}=\frac{M_x}{m}$ . in polar coordinate $y=r\sin\theta$

$\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta$

$=3k\int_{0}^{\pi}\sin\theta d\theta$

$=3k\big[-\cos\theta\big]_{0}^{\pi}$

$=3k\big[-\cos\pi+\cos 0\big]$

$=6k$

Then, $\bar{y}=\frac{M_x}{m}=\frac{2}{\pi}$

y-coordinate of center of mass is $\bar{x}=\frac{M_y}{m}$ . in polar coordinate $x=r\cos\theta$

$\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta$

$=3k\int_{0}^{\pi}\cos\theta d\theta$

$=3k\big[\sin\theta\big]_{0}^{\pi}$

$=3k\big[\sin\pi-\sin 0\big]$

$=0$

Then, $\bar{x}=\frac{M_y}{m}=0$

Hence center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

3 0

3 years ago