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Mice21 [21]
3 years ago
12

What is the length of side PQ in this figure?

Mathematics
2 answers:
balandron [24]3 years ago
7 0

Answer:

option: C

Step-by-step explanation:

To find the side PQ ; we need to first find the length of the given line segment which is perpendicular to side PR; let name it as QS.

Now as ΔQSR is an right angled triangle.

and the length of the side QR and SR is given , so using Pythagorean theorem we have

QR^{2}=SR^{2}+QS^{2}

5^{2}=3^{2}+QS^{2}\\ \\QS^{2}=5^{2}-3^{2}\\\\QS^{2}=25-3=16=4^2

⇒  QS=4

Now again ΔQSP is an right angled triangle; so using Pythagorean theorem in ΔQSP we have

PQ^{2}=QS^{2}+PS^{2}\\ \\PQ^{2}=4^{2}+6^{2}=16+36=52

This means QS=\sqrt{52}=2\sqrt{13}

Hence, the length of side PQ is \sqrt{52}=2\sqrt{13}.

Hence, option C is correct.




Elina [12.6K]3 years ago
5 0
I hope this helps you

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Solve 12^x^2+5x-4 = 12^2x+6
faust18 [17]

The solutions for ‘x’ are 2 and -5

<u>Step-by-step explanation:</u>

Given equation:

                    12^{x^{2}+5 x-4}=12^{2 x+6}

Since the base on both sides as ‘12’ are the same, we can write it as

                     x^{2}+5 x-4=2 x+6

                     x^{2}+5 x-2 x-4-6=0

                     x^{2}+3 x-10=0

Often, the value of x is easiest to solve by a x^{2}+b x+c=0 by factoring a square factor, setting each factor to zero, and then isolating each factor. Whereas sometimes the equation is too awkward or doesn't matter at all, or you just don't feel like factoring.

<u>The Quadratic Formula:</u> For a x^{2}+b x+c=0, the values of x which are the solutions of the equation are given by:

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

Where, a = 1, b = 3 and c = -10

                       x=\frac{-3 \pm \sqrt{(-3)^{2}-4(1)(-10)}}{2(1)}

                       x=\frac{-3 \pm \sqrt{9+40}}{2}

                       x=\frac{-3 \pm \sqrt{49}}{2}=\frac{-3 \pm 7}{2}

So, the solutions for ‘x’ are

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                       x=\frac{-3-7}{2}=\frac{-10}{2}=-5

The solutions for ‘x’ are 2 and -5

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3 years ago
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