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Arada [10]
3 years ago
5

Find the equation of the line which passes through the point (-4,12) and is perpendicular to the given line. Express your answer

in slope intercept form. The slope is -7
Mathematics
1 answer:
Vika [28.1K]3 years ago
8 0

Answer:

y = 1/7x + 88/7

Step-by-step explanation:

y = mx +b

since it is perpendicular you have to get the negative reciprocal of the slope which is 1/7

y = 1/7x + b then you would plug in your points

12 = 1/7(-4) + b and then solve for b

12 = -4/7 + b

+ 4/7

88/7 = b

y = 1/7x + 88/7

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Mr. Herkowitz owns a jewelry store. He marks up all merchandise 50 percent of
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Sebastian swam laps every day in the community swimming pool. He swam 45 minutes each day, 5 days each week, for 12 weeks. In th
lord [1]

Answer:

\text{Average rate}=\frac{\text{40 laps}}{\text{ hour}}

Step-by-step explanation:

Please consider the complete question.

Sebastian swan laps every day in the community swimming pool. He swam 45 minutes each day, 5 days each week, for 12 weeks. In that time, he swam 1800 laps. What was his average rate in laps per hour?    

Let us convert time taken into hours as:

\text{Days}=5\times 12=60

\text{Minutes}=60\times 45

1 hour = 60 minutes. To convert minutes into hours, we will divide total minutes by 60.

\text{Hours}=\frac{60\times 45}{60}=45

Now, we will divide 1800 by 45 to find average rate in laps per hour as:

\text{Average rate}=\frac{\text{1800 laps}}{\text{45 hours}}

\text{Average rate}=\frac{\text{40 laps}}{\text{ hour}}

Therefore, Sebastian's average rate is 40 laps per hour.

6 0
3 years ago
Write a standard form of a quadratic equation that has roots - 5/2 and 3.
lora16 [44]

Answer:

2x^2 - x - 15 = 0.

Step-by-step explanation:

In factor form this is

(x + 5/2)(x - 3) = 0

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x^2 - 1/2x - 15/2 = 0

Multiply through by 2:

2x^2 - x - 15 = 0.

6 0
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250 is equal to the ratio ____ to 100​
Ksivusya [100]

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25

Step-by-step explanation:

if you look at it, 250 is 1/4 of 1000 and 100 is 1/10 of 1000 so divide 250 by 10 and you get your answer.

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3 years ago
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