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Ivan
3 years ago
13

142=10x+8x+2. Please show steps as well.​

Mathematics
2 answers:
mel-nik [20]3 years ago
7 0

Answer:

x=7.78

Step-by-step explanation:

142=10x+8x+2

Move all numbers with the same variables to each side to separate them.

142-2=10x+8x

140=18x

Divide by 18.

x= 140/18

x=7.78

Alika [10]3 years ago
4 0
You cannot put the answer in decimal form bc there are no decimals in the equation therefore the answwer is x=70/9

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A tortoise walks 3 inches in 1 second. How many feet per second can the tortoise walk?
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0.25

Step-by-step explanation:

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3 years ago
A. -3<br> B. -4<br> C. -2 and 1<br> D. 0
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Read 2 more answers
What is the slope of the line that passes through the points (−10,−8) and (-8, -16) Write your answer in simplest form.
jarptica [38.1K]

Step-by-step explanation:

given,

point = ( -10, -8 ) and ( -8, -16 )

we know,

slope = y2 - y1 / x2 - x1

x1 , y1 = (-10 , -8 )

x2 , y2 = ( -8, -16 )

after inserting the values we got,

→ -16 - (-8 ) / -8 - (-10)

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therefore, slope of line is -4.

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7 0
3 years ago
Simplify expression x8y-26/x14y-5 × x-39y-21
koban [17]

The simplified form of the expression is x^{-45} y^{-42}

<h3>Simplifying an expression</h3>

From the question, we are to simplify the given expression

The given expression is

\frac{x^{8}y^{-26}  }{x^{14}y^{-5}  } \times x^{-39} y^{-21}

The expression can be simplified as shown below

\frac{x^{8}y^{-26}  }{x^{14}y^{-5}  } \times x^{-39} y^{-21}

\frac{x^{8} \times x^{-39} \times  y^{-26} \times  y^{-21}  }{x^{14} \times y^{-5}  }

\frac{x^{8+-39} \times  y^{-26+-21}  }{x^{14} \times y^{-5}  }

\frac{x^{8-39} \times  y^{-26-21}  }{x^{14} \times y^{-5}  }

\frac{x^{-31} \times  y^{-47}  }{x^{14} \times y^{-5}  }

Then,

\frac{x^{-31} }{x^{14} } \times \frac{ y^{-47} }{ y^{-5} }

x^{-31-14} \times y^{-47--5}

x^{-31-14} \times y^{-47+5}

x^{-45} \times y^{-42}

x^{-45} y^{-42}

Hence, the simplified form of the expression is x^{-45} y^{-42}

Learn more on Simplifying an expression here: brainly.com/question/2320607

#SPJ1

3 0
2 years ago
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