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blondinia [14]
2 years ago
10

Write the product in its simplest form: бу - бу?

Mathematics
1 answer:
weqwewe [10]2 years ago
4 0

Answer:

0

Step-by-step explanation:

This deals with simplification or other simple results.

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A cylinder has a 12-inch diameter and is 15 inches tall. It is filled to the top with water. A 6-inch-diameter ball is placed wi
Tems11 [23]

The volume of the cylinder is the space occupied by the cylinder. The volume of the water in the cylinder is 1583.36266 in³.

<h3>What is the volume of a cylinder?</h3>

The volume of the cylinder is the space occupied by the cylinder. It is calculated with the help of the formula,

\text{Volume of the cyllinder}= \pi r^2h

As it is given that the diameter of the cylinder is 12 inches while the height of the cylinder is 15 inches, also, it has a ball of the diameter of 6 inches placed inside the cylinder, therefore, the volume of the water that is inside the cylinder is the difference in the volume of the cylinder and the volume of the spherical ball.

\text{Volume of the cyllinder}= \pi r^2h = \dfrac{\pi}{4}d^2 h

Substitute the values,

\begin{aligned}\text{Volume of the cyllinder}&= \dfrac{\pi}{4}d^2 h\\& = \dfrac{\pi}{4} \times (12^2) \times 15\\ &=1696.46\rm\ in^3\end{aligned}

Now, the volume of the ball is equal to the volume of the sphere therefore, the volume of the ball can be written as,

\text{Volume of the sphere} = \dfrac{4}{3}\pi r^3

                                  = \dfrac{4}{3}\pi r^3\\\\=  \dfrac{4}{3}\times \pi \times (3^3)\\\\= 113.097\rm\ in^3

Further, the volume of the water that is inside the cylinder can be written as,

The volume of water =  Volume of the cylinder - Volume of the sphere

                                   = 1696.46 - 113.097

                                   = 1583.36266 in³

Hence, the volume of the water in the cylinder is 1583.36266 in³.

Learn more about Volume of the Cylinder:

brainly.com/question/1780981

5 0
2 years ago
Help I’ll give brainliest
ikadub [295]

Answer:

b

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Please help meee and show all steps please!!!ty
kolbaska11 [484]

Answer:

x = -4 and x = 5

Step-by-step explanation:

Since x^2 + 2x and 3x + 20 both equal to y, we know that the expressions equal to each other. We can write a new equation base on that.

x^2 + 2x = 3x + 20

Now we solve the equation.

x^2 + 2x = 3x + 20

x^2 - x = 20

x^2 - x - 20 = 0

(x + 4) (x - 5) = 0

x + 4 = 0 ; x -5 = 0

x = -4 ; x = 5

3 0
2 years ago
Find the domain of f(x) = x+5/ 2-x
lions [1.4K]
I don’t know I am stuck on thisss
8 0
2 years ago
Place the three functions in order from the fastest decreasing average rate of change to the slowest decreasing average rate of
victus00 [196]

Answer:

g(x), f(x) and h(x)

Step-by-step explanation:

Given

Interval: (0,3)

See attachment for functions f(x), g(x) and h(x)

Required

Order from fastest to slowest decreasing average rate of change

The average rate of change is calculated as:

Rate = \frac{f(b) - f(a)}{b - a}

In this case:

(a,b) = (0,3)

i.e.

a = 0\\b=3

For f(x)

f(x) = 16(\frac{1}{2})^x

Rate = \frac{f(b) - f(a)}{b - a}

Rate = \frac{f(3) - f(0)}{3 - 0}

Rate = \frac{f(3) - f(0)}{3}

Calculate f(3) and f(0)

f(x) = 16(\frac{1}{2})^x

f(3) = 16(\frac{1}{2})^3 = 16 * \frac{1}{8} = 2

f(0) = 16(\frac{1}{2})^0 = 16 * 1 = 16

So:

Rate = \frac{f(3) - f(0)}{3}

Rate = \frac{2 - 16}{3}

Rate = -\frac{14}{3}

For g(x)

Rate = \frac{g(b) - g(a)}{b - a}

Rate = \frac{g(3) - g(0)}{3 - 0}

Rate = \frac{g(3) - g(0)}{3}

From the table of g(x)

g(3) = 1

g(1) = 27

So:

Rate = \frac{1 - 27}{3}

Rate = -\frac{26}{3}

For h(x)

Rate = \frac{h(b) - h(a)}{b - a}

Rate = \frac{h(3) - h(0)}{3 - 0}

Rate = \frac{h(3) - h(0)}{3}

From the graph of h(x)

h(3) = -3

h(0) = 4

So:

Rate = \frac{-3 - 4}{3}

Rate = -\frac{7}{3}

So, the calculated rates of change are:

f(x) = -\frac{14}{3} = -4.67

g(x) = -\frac{26}{3} =-8.67

h(x) = -\frac{7}{3} =-2.33

By comparison:

From the fastest decreasing to slowest, the order is: <em>g(x), f(x) and h(x)</em>

4 0
3 years ago
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