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Marina86 [1]
3 years ago
11

H/j=15 solve for j Please answer

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
5 0
Hey!

Let's write the problem,
\frac{H}{j}=15
Multiply both sides by j in order to undo the fraction.
\frac{H}{j}j=15j
H=15j
Let's switch sides so that we end up with j on the left side.
15j=H
Divide both sides by 15.
\frac{15j}{15}=\frac{H}{15}

Our final answer would be,
j=\frac{H}{15}

Thanks!
-TetraFish
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The function f(x)=1+1.5ln(x+1) models the average number of free-throws a basketball player can make consecutively during practi
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Answer:

402 days.

Step-by-step explanation:

f(x)=1+1.5ln(x+1)

When f() = 10 we have:

10 = 1 + 1.5ln (x + 1)

ln (x + 1) = (10-1) / 1.5 = 6

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3 years ago
Match each set of points with the slope of the line formed by the ordered pairs.
Fynjy0 [20]

1) The slope of the point (0,0) and (2,1) is 1/2.

2) The slope of the point (0,-1) and (2,-5) is -2.

3) The slope of the point (0,7) and (2,4) is -3/2.

4) The points (-1,3) and (-1,-3) has No slope.

5) The slope of the point (1,-3) and (-1,-3) is zero.

<u>Step-by-step explanation:</u>

There are five set of points given in the question.

You need to find out the slope of the line formed by the points.

The formula used to find the slope of a line is given by,

Slope = \frac{y2-y1}{x2-x1}

1) The given points are (0,0) and (2,1)

(x1,y1) ⇒ (0,0)

(x2,y2) ⇒ (2,1)

Slope = \frac{1-0}{2-0}

⇒ 1/2 is the slope.

2) The given points are (0,-1) and (2,-5)

(x1,y1) ⇒ (0,-1)

(x2,y2) ⇒ (2,-5)

Slope = \frac{-5+1}{2-0}

⇒ \frac{-4}{2}

⇒ -2 is the slope.

3) The given points are (0,7) and (2,4)

(x1,y1) ⇒ (0,7)

(x2,y2) ⇒ (2,4)

Slope = \frac{4-7}{2-0}

⇒ -3/2 is the slope.

4) The given points are (-1,3) and (-1,-3)

(x1,y1) ⇒ (-1,3)

(x2,y2) ⇒ (-1,-3)

Slope = \frac{-3-3}{-1+1}

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If the denominator of a fraction is zero, the expression is not a legal fraction because it's overall value is undefined.

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(x2,y2) ⇒ (-1,-3)

Slope = \frac{-3+3}{-1-1}

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Select the correct answer from the drop-down menu.
ELEN [110]

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Taking common term xy outside in the numerator.

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