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mestny [16]
3 years ago
9

What is equivalent to 5^3 ( Choose all that apply)

Mathematics
1 answer:
Tema [17]3 years ago
8 0

Answer:

A, and D

Step-by-step explanation:

A works because when you multiply like bases, you add exponents, so we have...

  5^7*5^-4 = 5^(7+(-4)) = 5^(7 -4) = 5^3

B  doesn't work because when you divide like bases, you subtract the exponent of the denominator from the exponent of the denominator...

  5^12/5^4 = 5^(12 - 4) = 5^8

C doesn't work because there is addition.  

 5 + 5² = 5 + 25 = 30, and 5³ = 125

D works because 5^0 = 1, so we have

  5^0(5^3) = 1(5^3) = 5^3

E doesn't work because there is subtraction...

  5^3 - 5^0 = 5^3 - 1 = 124

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What are the solutions of x^2-2x+5=0
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The solution of x^{2}-2 x+5=0 are 1 + 2i and 1 – 2i

<u>Solution:</u>

Given, equation is x^{2}-2 x+5=0

We have to find the roots of the given quadratic equation

Now, let us use the quadratic formula

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}  --- (1)

<em><u>Let us determine the nature of roots:</u></em>

Here in x^{2}-2 x+5=0 a = 1 ; b = -2 ; c = 5

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Now plug in values in eqn 1, we get,

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On solving we get,

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x=\frac{2 \pm \sqrt{-16}}{2}

x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}

we know that square root of -1 is "i" which is a complex number

\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}

Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i

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