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3 years ago

8

Find the volume of a cylinder that has the following dimensions. Do not round your answer. (Use 3.14 for π.)

1 answer:

denpristay [2]3 years ago

4 0

Ok this is easy so since the formula for a cylinder is pie(r to the 2nd power) times h it’s like this——-

3.14(5 to the 2nd power) 0.25
3.14(25)0.25
78.5(0.25)
=19.625

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X to the power of 2+ 2X plus 3X to the power of 2+ 4X subtract 2X to the power of 2+ X

eimsori [14]

Answer:

4x+12

Step-by-step explanation:

$2x + 2 \times { }^{2} + 6x + 12 x ^{2} - 4x + - 2x ^{2}$

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3 years ago

What is this. Please help!!

Mekhanik [1.2K]

Answer: 8x+24

Step-by-step explanation:

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3 years ago

Choose the option that best answers the question.The points A(0, 0), B(0, 4a – 5) and C(2a + 1, 2a + 6) form a triangle. If Angl

Sunny_sXe [5.5K]

Answer:

The correct option is 1.

Step-by-step explanation:

Given information: The coordinates of a right angled triangle ABC are A(0, 0), B(0, 4a – 5) and C(2a + 1, 2a + 6). Angle ABC = 90°.

It means AB and BC are legs of the right angled triangle ABC.

Side AB lies on the y-axis because the x-coordinate of both A and B is 0.

Two legs are perpendicular to each other. So, BC must be parallel to x-axis and the y-coordinate of both B and C is must be same.

$4a-5=2a+6$

$4a-2a=5+6$

$2a=11$

Divide both sides by 2.

$a=\frac{11}{2}$

The value of a is 2. So the coordinates of triangle ABC are

$B(0,4a-5)=B(0,4(\frac{11}{2})-5)\Rightarrow B(0,17)$

$C(2a+1,2a+6)=C(2(\frac{11}{2})+1,2(\frac{11}{2})+6)\Rightarrow C(12,17)$

The area of a triangle is

$Area=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

The area of triangle ABC is

$Area=\frac{1}{2}|0(17-17)+0(17-0)+12(0-17)|$

$Area=\frac{1}{2}|12(-17)|$

$Area=\frac{1}{2}|-204|$

$Area=\frac{1}{2}(204)$

$Area=102$

The area of triangle ABC is 102. Therefore the correct option is 1.

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3 years ago

Will give brainliest please help

Romashka [77]

Answer 2/5. It says there are 5 people and 3 of them have done their hw and 2 have not cause 5-3=2.

8 0

3 years ago

2.<br> Which of the functions below is an inverse of the quadratic function f(x) = x2 - 3 ?

mezya [45]

Answer:

the inverse of the quadratic function $f(x) = x^2 - 3$ is $\mathbf{f^{-1}(x)=\pm \sqrt{x+3}}$

Step-by-step explanation:

We need to find inverse of the quadratic function $f(x) = x^2 - 3$

Let

$y=x^2-3$

Replace x and y

$x=y^2-3$

Now, we will solve for y

Adding 3 on both sides

$x+3=y^2-3+3$

$x+3=y^2$

Taking square root on both sides

$\sqrt{y^2}=\sqrt{x+3}\\y=\pm \sqrt{x+3}$

Now replace y with $f^{-1}(x)$

$f^{-1}(x)=\pm \sqrt{x+3}$

So, the inverse of the quadratic function $f(x) = x^2 - 3$ is $\mathbf{f^{-1}(x)=\pm \sqrt{x+3}}$

3 0

2 years ago