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emmasim [6.3K]
3 years ago
7

Please help me thanks

Mathematics
2 answers:
astra-53 [7]3 years ago
4 0

Answer:

C

Step-by-step explanation:

X is 10 so it is c

Hope this works

Ulleksa [173]3 years ago
3 0

<em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em>

<em>The</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>of</em><em> </em><em>option</em><em> </em><em>C</em>

<em>so</em><em> </em><em>if</em><em> </em><em>we</em><em> </em><em>have</em><em> </em><em>to</em><em> </em><em>solve</em><em> </em><em>it</em><em> </em><em>then</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>be</em><em>:</em>

<em>5</em><em>x</em><em>+</em><em>1</em><em>3</em><em>x</em><em>=</em><em>1</em><em>8</em><em>0</em>

<em>or</em><em>,</em><em>1</em><em>8</em><em>x</em><em>=</em><em>1</em><em>8</em><em>0</em>

<em>or</em><em>,</em><em>X=</em><em>1</em><em>8</em><em>0</em><em>/</em><em>1</em><em>8</em>

<em>X=</em><em>1</em><em>0</em>

<em>hope</em><em> </em><em>it</em><em> </em><em>helps</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

<em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em>

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C. The first is a solution, but the second is not.

5x - y/3 = 13 ; (2,-9)
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5(2) - (-9/3) = 13 
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Write the following expression in standard for a + bi<br><br><img src="https://tex.z-dn.net/?f=%5Cfrac%7B6-i%7D%7B1%2Bi%7D" id="
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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

                       x=\frac{-3+7}{2}=\frac{4}{2}=2

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

The solutions for ‘x’ are 2 and -5

6 0
3 years ago
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