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den301095 [7]
2 years ago
6

Find the length of the third side. If necessary, round to the nearest tenth. 5 and 9

Mathematics
2 answers:
Vlad1618 [11]2 years ago
7 0

Answer:

is it a right triangle?

Step-by-step explanation:

expeople1 [14]2 years ago
6 0

Answer:

1. Is it a right triangle?

2. Are you looking for the hypotenuse?

Step-by-step explanation:

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Find the equivalent exponential expression (7^2)^3
atroni [7]

Answer: 117649 is your answer

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A square with a side length of 5 inches has a area of 25 in<br><br> true or false
Goryan [66]

Answer:

TRUE

Step-by-step explanation:

5 x 5 = 25 in²

6 0
3 years ago
Expand and simplify 5(3x+2)-2(4x-1)
Rama09 [41]

Answer:

15x+10-8x+2

7x+12

Step-by-step explanation:

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identify the postulate or theorem that proves the triangles congruent. AAS, SAA, ASA, The triangles cannot be proven congruent
nirvana33 [79]
AAS and SAA don't prove congruence. 
You can have two triangles that have AAS or SAA and are not congruent.

From this picture, these triangles are congruent by ASA .
4 0
3 years ago
​Find all roots: x^3 + 7x^2 + 12x = 0 <br> Show all work and check your answer.
Aliun [14]

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

<u>Solution:</u>

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}

<em><u>Therefore, the two roots are:</u></em>

\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
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