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nekit [7.7K]
3 years ago
12

When dividing fractions, the ________ term becomes the reciprocal.

Mathematics
2 answers:
Aleks04 [339]3 years ago
5 0

Answer: Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction. The first step to dividing fractions is to find the reciprocal (reverse the numerator and denominator) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators.

Step-by-step explanation:

Gekata [30.6K]3 years ago
3 0

Answer:

i think its the first...

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Find the perimeter (add all sides) of a square when side a= - 41x²
Dafna11 [192]

Answer:

-164x^2

Step-by-step explanation:

The Perimeter of a square is 4s where s is the length of 1 side.

So P=4(-41x^2) 41*4=164 ----> P=-164x^2

5 0
2 years ago
What is four and five eighths as a improper fraction and a mixed number
Afina-wow [57]
4 5/8 as an improper fraction is 37/8
——> 8*4=32 32+5=37
As a mixed number it is the same I think. 5 can’t decide into 8 so it stays 4 5/8.

Hopefully this helps...
5 0
3 years ago
Find the coordinates of the turning point for the equation of the graph below y=x^2+14x+45
USPshnik [31]

Answer:

(- 7, - 4 )

Step-by-step explanation:

Given a quadratic in standard form

y = ax² + bx + c ( a ≠ 0 )

Then the x- coordinate of the turning point is

x = - \frac{b}{2a}

y = x² + 14x + 45 ← is in standard form

with a = 1, b = 14 , then

x = - \frac{14}{2} = - 7

Substitute x = - 7 into the equation and evaluate for y

y = (- 7)² + 14(- 7) + 45 = 49 - 98 + 45 = - 4

coordinates of turning point = (- 7, - 4 )

4 0
2 years ago
Check all of the proper units of weight in the metric system.
MissTica
Gram and milligram are in the metric system and are for weight. Pound is the United States system and kiloliter measures liquids
4 0
3 years ago
From first principles, find the indicated derivatives​
LenaWriter [7]

By definition of the derivative,

\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{(s + h)^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}

\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{s^3+3s^2h+3sh^2+h^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}

\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\frac{3s^2h+3sh^2+h^3}2}{h}

\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 \frac{3s^2h+3sh^2+h^3}{h}

\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 (3s^2+3sh+h^2)

\displaystyle\frac{dr}{ds} = \frac{3s^2}2

6 0
2 years ago
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