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8_murik_8 [283]
3 years ago
14

What value of x makes the equation true?

Mathematics
2 answers:
jek_recluse [69]3 years ago
8 0

Answer:

A.) x = -4

Step-by-step explanation:

<em><u>1. Simplify the expression</u></em>

1.4x + 2.6x = -16

Simplify the arithmetic:

4x = -16

<u><em>2. Isolate the x</em></u>

4x = -16

Divide both sides by 4:

\frac{4x}{4} = \frac{-16}{4}

Simplify the fraction:

x = \frac{-16}{4}

Find the greatest common factor of the numerator and denominator:

x = \frac{-4 * 4}{1 * 4}

Factor out and cancel the greatest common factor:

x = -4

A.) x = -4

gtnhenbr [62]3 years ago
4 0

Answer:

1.4x + 2.6x =  - 16 \\ 4x =  - 16 \\ x =  \frac{ - 16}{4}  \\ x =  - 4 \\ thank \: you

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Answer this question please and thank you
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Answer:

\frac{1}{729}

Step-by-step explanation:

1) Use the division distribution property: (x/y)^a = x^a/y^a.

\frac{1}{9^{3} }

2) Simplify 9³ to 729.

\frac{1}{729}

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8 0
3 years ago
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What does 1/2 (40x60) =
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40*60=2400
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Write the given trinomial if possible as a square of a binomial or as an expression opposite to a square of a binomial: 15ab-9a^
MrMuchimi

Answer:

- \bigg(3a -  \frac{5}{2} b) \bigg)^{2}

Step-by-step explanation:

15ab-9a^2-6  \frac{1}{4} b^2 \\  \\  = 15ab-(3a)^2-\frac{6 \times 4 + 1}{4} b^2 \\  \\ = 15ab-(3a)^2-\frac{24+ 1}{4} b^2 \\  \\ = 15ab-(3a)^2-\frac{25}{4} b^2 \\  \\   = 15ab- (3a)^2- \bigg(\frac{5}{2} b \bigg)^2  \\  \\ =  -  \{ - 15ab + (3a)^2 +  \bigg(\frac{5}{2} b \bigg)^2  \} \\  \\ =  -  \{ (3a)^2 +  \bigg(\frac{5}{2} b \bigg)^2 - 15ab  \} \\  \\  =   - \bigg(3a -  \frac{5}{2} b \bigg)^{2}

3 0
3 years ago
Find the midpoint of the segment having endpoints (0 ,1/8) and (-4/5 ,0)
kvasek [131]

Answer:

Mid point of the given end points = (\frac{-2}{5},\frac{1}{16} )

Step-by-step explanation:

Given the end points of the line segment are :

(0, \frac{1}{8}) & (\frac{-4}{5},0)

We will use the mid point formula when two points are: (x₁,y₁) & (x₂,y₂)

mid point = (\frac{x_{1} +x_{2} }{2},\frac{y_{1} +y_{2} }{2} )

now put the value of x₁ = 0

x₂ = \frac{-4}{5}

y₁ = \frac{1}{8}

y₂ = 0

mid point = (\frac{0-\frac{4}{5} }{2},\frac{\frac{1}{8}+0 }{2})

                = (\frac{-2}{5},\frac{1}{16} )

That's the final answer.

3 0
3 years ago
(8v)^-1 writer with positive exponent and then simplify.
IgorLugansk [536]
(8v)^{-1}= \cfrac{1}{8v^1} = \cfrac{1}{8v}
6 0
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