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Delicious77 [7]
3 years ago
9

PROBLEM. 11.1: Rewrite These! Rewrite each quotient as a sum or a difference.

Mathematics
1 answer:
Jet001 [13]3 years ago
3 0

Answer:

1. (4x-10)/2

   4x/2 - 10/2

    2x - 5

2. (1-50x)/-2

   1/-2 -50x/-2

   25x - 1/2

3. 5(x + 10)/25

   (5x + 50)/25

   5x/25 + 50/25

   1/5 x + 2

4.(-0.2x + 5)/2​

   -0.2x/2 + 5/2

   -0.1x + 2.5

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Answer:

y=\frac{1}{5}x+\frac{12}{5}

Step-by-step explanation:

<u><em>Complete Question:</em></u> What is the equation of line that is parallel to the line y=\frac{1}{5}+4 and passes through (-2,2).

Let\ the\ equation\ of\ line\ is\ y=mx+c\\\\Since\ it\ is\ parallel\ to\ the\ line\ y=\frac{1}{5}x+4,\ the\ slope\ of\ both\ the\ lines\ will\ be\ same.\\\\Slope\ of\ y=\frac{1}{5}x+4\ is=\frac{1}{5}\\\\m=\frac{1}{5}\\\\Equation: y=\frac{1}{5}x+c\\\\It\ passes\ through\ (-2,2),\ This\ point\ will\ satisfy\ the\ equation.\\\\2=\frac{1}{5}\times (-2)+c\\\\c=2+\frac{2}{5}\\\\c=\frac{12}{5}\\\\The\ Equation\ is: y=\frac{1}{5}x+\frac{12}{5}

7 0
3 years ago
When building the Channel, the English and French each started drilling on opposite sides of the English Channel. The two sectio
Alekssandra [29.7K]

Suppose the English could drill the Channel in 6.2 years, then they drill with speed \dfrac{L}{6.2} km per year, where L is the length of the English Channel. If the French could drill in 5.8 years, then their speed of drillling is \dfrac{L}{5.8} km per year.

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\dfrac{L}{6.2} +\dfrac{L}{5.8} =\dfrac{5.8L+6.2L}{6.2\cdot 5.8}=\dfrac{12L}{35.96}=\dfrac{3L}{8.99}.

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jenyasd209 [6]

Answer:

Step-by-step explanation:

1. List out all data from least to greatest

2. Find the median (the middle number)

3. Find the median of the lower part of data (the numbers lesser in value compared to the median; do not include the median while finding this)

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Please respond in detail
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Answer:

No.

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As x increases 2^x will grow faster than 5 x^2. This is because the x in 2^x is an exponent and its graph grows very steeply compared with 5x^2 , as x increases.

For example when x = 10

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When x = 11:

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