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3 years ago

Help me...............

Mathematics

1 answer:

True [87]3 years ago

3 0

Answer:

Brainliestgive o020201000

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Guys, please please help me

agasfer [191]

Answer:

508.68

Step-by-step explanation:

Upper and lower surfaces of the cylinder:

$\pi r^{2} =9\pi \\S=2*9\pi =18\pi$

Side area of the cylinder:

$S=sh=\pi dh=6\pi *3*8=144\pi$

The total area:

$S=114\pi +18\pi =162\pi =162*3.14=508.68$

5 0

3 years ago

How can you use an algebraic expression on a coordinate plane?

nydimaria [60]

Putting it in the middle sorry if it isnt right i never was really good at these but i tried.

4 0

3 years ago

What is the value of the expression? 8 1/2-2+4 3/4

kodGreya [7K]

11 1/4 or 11.25 is the answer

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3 years ago

Find all roots: x^3 + 7x^2 + 12x = 0 Show all work and check your answer.

Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

Solution:

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

$x^{3}+7 x^{2}+12 x=0$

$x\left(x^{2}+7 x+12\right)=0$ ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

$x^{2}+7 x+12=0$

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

$\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}$

Therefore, the two roots are:

$\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}$

And,

$\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}$

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0

3 years ago

Enrique was asked to solve the equation His steps are shown.

valkas [14]

there is nothing Shown on the steps

6 0

3 years ago

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