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aleksandr82 [10.1K]
2 years ago
7

Which expression is equivalent to

Mathematics
2 answers:
laila [671]2 years ago
4 0

Answer:

-8

Step-by-step explanation:

xxMikexx [17]2 years ago
3 0

Answer:

-8

Step-by-step explanation:

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Hello can you please help me posted picture of question
jarptica [38.1K]
4x^6+2x^5-2x+8+2x^8+4x+2=
2x^8+4x^6+2x^5+(4-2)x+10=
2x^8+4x^6+2x^5+2x+10

Answer: Option B: 2x^8+4x^6+2x^5+2x+10
7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx-3%7D%7Bx%2B2%7D%20%3D%5Cfrac%7B1%7D%7B5%7D" id="TexFormula1" title="\frac{x-3}{x+2
Shkiper50 [21]

Answer:

x=17/4

Step-by-step explanation:

(x-3)/(x+2)=1/5

cross product

1(x+2)=5(x-3)

x+2=5x-15

5x-x-15=2

4x-15=2

4x=2+15

4x=17

x=17/4

3 0
3 years ago
What is 2∙ 2∙ 2∙ 2∙ 2∙ 2∙ 2∙ 2∙ 2 in exponential form?
Novay_Z [31]

Answer: 0.0001408

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
What is the value of (3/10)3
Elenna [48]

Answer:

\frac{9}{10}

Step-by-step explanation:

In this question, one is asked to multiply a fraction by a whole number, first model the question.

\frac{3}{10}(3)

In a situation like this, one multiplies the whole number by the numerator (number) over the fraction.

\frac{3*3}{10}

Simplify,

\frac{9}{10}

A common error that people make is that they will multiply both the numerator and denominator by the whole number, this is the same as multiplying the answer by (1), thus it is incorrect.

6 0
2 years ago
Find all the zeros of the equation<br> -3x^4 + 27x^2 + 1200 = 0
guajiro [1.7K]

Answer:

\displaystyle x=-5,\ x=5,\ x=4i,\ x=-4i

Step-by-step explanation:

Biquadratic Equation

It's a fourth-degree equation where the terms of degree 1 and 3 are missing. It can be solved for the variable squared as if it was a second-degree equation, and then take the square root of the results

Our equation is

\displaystyle -3x^4+27x^2+1200=0

If we call y=x^2, our equation becomes a second-degree equation

\displaystyle -3y^2+27y+1200=0

Dividing by -3

\displaystyle y^2-9y-400=0

Factoring

\displaystyle (y-25)(y+16)=0

It leads to these solutions

\displaystyle y=25\ ,\ y=-16

Taking back the change of variable, we have for the first solution

\displaystyle x^2=25\Rightarrow x=-5,x=5

Now for the second solution, we get imaginary (complex) values

\displaystyle x^2=-16\Rightarrow x=4i,\ x=-4i

Summarizing, the four solutions for x are

\displaystyle x=-5,\ x=5,\ x=4i,\ x=-4i

5 0
3 years ago
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