Ask question

Ask question

All categories

scoundrel [369]

2 years ago

10

If given the edge of a cube blank to determine the volume

1 answer:

tester [92]2 years ago

4 0

You would cube the edge (raise to an exponent of three)

You might be interested in

Find the value of x in the following figure

tiny-mole [99]

Answer:

8

Step-by-step explanation:

3 0

3 years ago

A circle with radius 3 has a sector with a central angle of 1/9 pi radians

marissa [1.9K]

Complete question:

A circle with radius 3 has a sector with a central angle of 1/9 pi radians

what is the area of the sector?

Answer:

The area of the sector = $\frac{\pi}{2}$ square units

Step-by-step explanation:

To find the area of the sector of a circle, let's use the formula:

$A = \frac{1}{2} r^2 \theta$

Where, A = area

r = radius = 3

$\theta = \frac{1}{9}\pi$

Substituting values in the formula, we have:

$A = \frac{1}{2}*3^2* \frac{1}{9}\pi$

$A = \frac{1}{2}*9* \frac{1}{9}\pi$

$A = 4.5 * \frac{1}{9}\pi$

$A = \frac{\pi}{2}$

The area of the sector = $\frac{\pi}{2}$ square units

8 0

3 years ago

I have 30 points to give away. One person answers this question and takes 10 points. If another person takes 10 more points, how

Ghella [55]

Answer:

10 points

Step-by-step explanation:

plz brainleist

8 0

3 years ago

Jonathan purchased 30 markers for $5.40. What is the price per marker?

madreJ [45]

Answer: 0.18 is your answer

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

Elena L [17]

Answer:

The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, a = $\sqrt{(3-0)^{2}+(4-0)^{2}}$

Or, a = $\sqrt{9+16}$

Or, a = $\sqrt{25}$

∴ a = 5 unit

Similarly

The measure of length BC = b = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, b = $\sqrt{(3-3)^{2}+(2-4)^{2}}$

Or, a = $\sqrt{0+4}$

Or, b = $\sqrt{4}$

∴ b = 2 unit

And

So, The measure of length CA = c = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, c = $\sqrt{(3-0)^{2}+(2-0)^{2}}$

Or, c = $\sqrt{9+4}$

Or, c = $\sqrt{13}$

∴ c = $\sqrt{13}$ unit

Now, area of Triangle written as , from Heron's formula

A = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

and s = $\frac{a+b+c}{2}$

I.e s = $\frac{5+2+\sqrt{13}}{2}$

Or. s = $\frac{7+\sqrt{13}}{2}$

So, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}$

Or, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}$

Or, A = $\frac{3}{2}$ × $\sqrt{1+\sqrt{13} }$

∴ Area of triangle = 1.5 $\sqrt{4.6}$

Hence The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$ Answer

7 0

2 years ago

Read 2 more answers