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alukav5142 [94]
2 years ago
10

Which expression below has the same value?

Mathematics
2 answers:
UNO [17]2 years ago
3 0

Answer:A

Step-by-step explanation:

Art [367]2 years ago
3 0

Answer:

15/8

Step-by-step explanation:

You will want all numbers to have the same common denominator, then you can add them together.

1 = 8/8

1/2 = 4/8

1/4 = 2/8

1/8 = 1/8

1+\frac{1}{2} +\frac{1}{4} +\frac{1}{8} \\\frac{8}{8} +\frac{4}{8} +\frac{2}{8} +\frac{1}{8}

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4 0
3 years ago
1/6=?/18 this is for college basic math fractions
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3 years ago
Read 2 more answers
What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve.
lapo4ka [179]

Consider the equation x^6+6x^3+5=0.

First, you can use the substitution t=x^3, then x^6=(x^3)^2=t^2 and equation becomes t^2+6t+5=0. This equation is quadratic, so

D=6^2-4\cdot 5\cdot 1=36-20=16=4^2,\ \sqrt{D}=4,\\ \\ t_{1,2}=\dfrac{-6\pm 4}{2} =-5,-1.

Then you can factor this equation:

(t+5)(t+1)=0.

Use the made substitution again:

(x^3+5)(x^3+1)=0.

You have in each brackets the expression like a^3+b^3 that is equal to (a+b)(a^2-ab+b^2). Thus,

x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)

and the equation is

(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25})(x+1)(x^2-x+1)=0.

Here x_1=-\sqrt[3]{5} , x_2=-1 and you can sheck whether quadratic trinomials have real roots:

1. D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}.

2. D_2=(-1)^2-4\cdot 1=1-4=-3.

This means that quadratic trinomials don't have real roots.

Answer: x_1=-\sqrt[3]{5} , x_2=-1

If you need complex roots, then

x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2}   ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}.

6 0
3 years ago
What is the slope of the line whose equation is 12=4x-6y?
jek_recluse [69]

Answer:

4x-6y-12=0 than it passes through the origin so.distance equals /ax+by+c/all over a^2+ b^2= 4(0)+6(0)-12/4^2+6^2= -12/16+36=12/52=0.23077

4 0
3 years ago
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