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3 years ago

5

What is the area of this triangle? (Image Attached)

2 answers:

marysya [2.9K]3 years ago

7 0

Answer:

Area of triangle is approximately 56.10 square inches.

Step-by-step explanation:

We are given a triangle, its two sides and angle between them.

We will use the side angle side formula to find the area of a triangle.

Area= $\dfrac{1}{2}\times 14.2\times 8\times sin99$

= $\dfrac{1}{2}\times 14.2\times 8\times 0.99$

= 56.10 square inches

Hence, Area of triangle is approximately 56.10 square inches.

Ivahew [28]3 years ago

6 0

56.1
1/2(14.2)(8)sin(99)=

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a cone with a radius of 3” has a total area of 24pie square inches. Find the volume of the cone. 12pie, 24 pie or 15 pie

Ratling [72]

Answer:

$12\pi (inches)^{2}$

Step-by-step explanation:

Given:

radius of cone = 3 inch

Total area of cone = $24\pi (inch)^{2}$

Find the volume of the cone?

We know the area of cone = $\pi r(r+s)$ -------(1)

Where r = radius

Ans s = side of the cone

Put the area value in equation 1

$24\pi=\pi r(r+s)$ -------(1)

$\frac{24\pi }{\pi r} =r+s$

$\frac{24}{ r}-r =s$

Put r value in above equation.

$s =\frac{24}{ 3}-3$

$s=8-3$

$s=5$

The side s = 5 inches

We know the side of the cone formula

$s^{2}=r^{2}+ h^{2}$

$h^{2}=s^{2}- r^{2}$

Put r and s value in above equation.

$h^{2}=5^{2}- 3^{2}$

$r^{2} = 25-9$

$r^{2} = 16$

$r = 4$

The volume of cone is

$V = \frac{1}{3} \pi r^{2} h$

Put r and h value in above equation.

$V = \frac{1}{3} \pi (3)^{2}\times 4$

$V = 12\pi (inches)^{2}$

The Volume of the cone is $12\pi (inches)^{2}$

7 0

2 years ago

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:

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

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

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3 years ago

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Step2247 [10]

7.3 also written as 7.30

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2 years ago

Which equation describes the relationship between the variables in the table below?

Dafna11 [192]

The answer would be D praise our lord zucc

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2 years ago

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How many five card hands can<br> be dealt from a standard deck of<br> 52 cards?

Law Incorporation [45]

Answer:

2

Step-by-step explanation:

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3 years ago