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Olegator [25]
3 years ago
5

What is the first step in solving the quadratic equation x2 = StartFraction 9 Over 16 EndFraction? Take the square root of both

sides of the equation. Add StartFraction 9 Over 16 EndFraction to both sides of the equation. Square both sides of the equation. Subtract StartFraction 9 Over 16 EndFraction from both sides of the equation.
Mathematics
3 answers:
stepan [7]3 years ago
7 0

Answer:

I'm gonna make this simple the answer is C on ed!!

balu736 [363]3 years ago
5 0

Answer:

The first step is

Take the square root of both sides of the equation

Step-by-step explanation:

we have

x^{2} =\frac{9}{16}

Solve for x

step 1

Take square root both sides

x=(+/-)\frac{3}{4}

the solutions are

x=+\frac{3}{4} and  x=-\frac{3}{4}

therefore

The first step is

Take the square root of both sides of the equation

lyra2 years ago
0 0

The second part might be "Take the square root of both sides"

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One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times sm
sergey [27]

One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle. Then the lines are parallel

<h3><u>Solution:</u></h3>

Given that, One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle.  

We have to prove that the lines are parallel.

If they are parallel, sum of the described angles should be equal to 180 as they are same side exterior angles.

Now, the 1st angle will be 1/6 of right angle is given as:

\begin{array}{l}{\rightarrow 1^{\text {st }} \text { angle }=\frac{1}{6} \times 90} \\\\ {\rightarrow 1^{\text {st }} \text { angle }=15 \text { degrees }}\end{array}

And now, 15 degrees is 11 times smaller than the other  

Then other angle = 11 times of 15 degrees

\text {Other angle }=11 \times 15=165 \text { degrees }

Now, sum of angles = 15 + 165 = 180 degrees.

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