Find the area of the shaded region 6cm 150°

Find the of the line

Umnica [9.8K]

Answer:

D

Step-by-step explanation:

Using the Slope Equation

Pick two points on the line and determine their coordinates.

Determine the difference in y-coordinates of these two points (rise).

Determine the difference in x-coordinates for these two points (run).

Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope

7 0

3 years ago

A rectangular prism has a top face with an area of 20 ft2 and a height of 5 ft. What is the volume of this rectangular prism? Th

kkurt [141]

Answer:

the volume of the rectangular prism is 100 ft³

Step-by-step explanation:

Given;

area of top face of a rectangular prism, A = 20 ft²

height of the rectangular prism, h = 5 ft

Volume of the rectangular prism = Area of top face x height of the rectangular prism

Volume of the rectangular prism = A x h

Volume of the rectangular prism = 20 ft² x 5 ft

= 100 ft³

Therefore, the volume of the rectangular prism is 100 ft³

6 0

3 years ago

Given The following Graph: a.)Evaluate f(0). b.) Solve for f(x) = -3

dlinn [17]

Answer:

see explanation

Step-by-step explanation:

a

f(0) means find the value of y when x = 0

That is f(0) = 1 ← the point (0, 1) on the graph

b

When f(x) = - 3 means what are the values of x corresponding to y = - 3

From the graph when y = - 3 there are 2 corresponding values of x, that is

x = - 2 or x = + 2

The solution to f(x) = - 3 is x = ± 2

3 0

4 years ago

The surface area of a cube is found using the formula sa=6s^2, where s is theside length if the surface area of a cube is 294 sq

aleksley [76]

6s^2 = 294
Solve for s

Solve for s:
6 s^2 = 294
Divide both sides by 6:
s^2 = 49
Take the square root of both sides:
Answer: s = 7 m

8 0

3 years ago

Read 2 more answers

A ball of radius 15 has a round hole of radius 5 drilled through its center. Find the volume of the resulting solid. Hint: The u

OverLord2011 [107]

Answer:

The volume of the ball with the drilled hole is:

$\displaystyle\frac{8000\pi\sqrt{2}}{3}$

Step-by-step explanation:

See attached a sketch of the region that is revolved about the y-axis to produce the upper half of the ball. Notice the function y is the equation of a circle centered at the origin with radius 15:

$x^2+y^2=15^2\to y=\sqrt{225-x^2}$

Then we set the integral for the volume by using shell method:

$\displaystyle\int_5^{15}2\pi x\sqrt{225-x^2}dx$

That can be solved by substitution:

$u=225-x^2\to du=-2xdx$

The limits of integration also change:

For x=5: $u=225-5^2=200$

For x=15: $u=225-15^2=0$

So the integral becomes:

$\displaystyle -\int_{200}^{0}\pi \sqrt{u}du$

If we flip the limits we also get rid of the minus in front, and writing the root as an exponent we get:

$\displaystyle \int_{0}^{200}\pi u^{1/2}du$

Then applying the basic rule we get:

$\displaystyle\frac{2\pi}{3}u^{3/2}\Bigg|_0^{200}=\frac{2\pi(200\sqrt{200})}{3}=\frac{400\pi(10)\sqrt{2}}{3}=\frac{4000\pi\sqrt{2}}{3}$

Since that is just half of the solid, we multiply by 2 to get the complete volume:

$\displaystyle\frac{2\cdot4000\pi\sqrt{2}}{3}$

$=\displaystyle\frac{8000\pi\sqrt{2}}{3}$

5 0

4 years ago