All categories

3 years ago

Write the ratio 15 inches to 9 feet as a fraction

1 answer:

6 0

We know that 1 ft= 12 in.

9 ft* (12 in/ 1 ft)= 108 in.

15 inches to 9 feet
= 15 in: 9 ft
= 15 in: 108 in (because 9 ft= 108 in)
= 15/108
= (15/3) / (108/3)
= 5/36

The final answer is 5/36~

You might be interested in

What is 95 divided by 5?

N76 [4]

95 divided by 5 is 19. You can count by fives until you get to 95.

3 0

2 years ago

Read 2 more answers

1. What is the radius (1 Point) 11 cm

mr_godi [17]

Given that we need to determine the radius of the circle.

Radius:

By definition of radius of circle, the radius is the length of the line which is drawn from the center of the circle to any point on the circle.

From the figure, we need to determine the radius of the circle.

As, we can see that, the distance from the center of the circle to the point on the circle is 11 cm.

Since, we know that, the radius is the distance from the center of the circle to the point on the circle then, the radius of the given circle is 11 cm

Thus, radius of the circle is 11 cm.

3 0

3 years ago

Which expression is equivalent to 4(7 + 8)? A) 11+8 B) 11+12 C) 28+12 D) 28+32

andreev551 [17]

To do this we take the outlier off the parenthesis (the 4) and multiply it by the two numbers inside the parenthesis...

4*7 + 4*8

28+32

7 0

3 years ago

Read 2 more answers

If the cone itself is 7 inches long, and the radius of the cone is 2 inches, what is the entire volume of ice cream in the cone

stepladder [879]

So if you use the volume of 7 inch and 2 inch your value is 14 volume

4 0

2 years ago

Read 2 more answers

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$

3 0

1 year ago