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2 years ago

Which number can each term of the equation be multiplied by to eliminate the fractions before solving?

Mathematics

1 answer:

Lerok [7]2 years ago

7 0

To eliminate the fraction, we multiply by 4

<h3>How to eliminate the fractions?</h3>

The equation is given as:

-3/4m-1/2=2+1/4m

Rewrite properly as:

$-\frac{3}{4}m - \frac 12 = 2 + \frac 14m$

The denominators of the above fractions are 2 and 4

The LCM of 2 and 4 is 4

So, we multiply through by 4.

This gives

-3m - 2 = 8 + m

Hence, to eliminate the fraction, we multiply by 4

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Decimals between 10 and 15 with an interval of 0.75

IgorC [24]

10.75
11.50
11.80
12.55
13.30
14.05
14.80
15.55

3 0

3 years ago

0.2(10-5c)=5c-16 halp

iogann1982 [59]

Answer:

c= 3

Step-by-step explanation:

0.2(10 -5c)= 5c -16

Expand:

0.2(10) +0.2(-5c)= 5c -16

2 -c= 5c -16

Bring all c terms to 1 side, constant to the other:

5c +c= 2 +16

Simplify:

6c= 18

Divide by 6 on both sides:

c= 18 ÷6

c= 3

5 0

3 years ago

Read 2 more answers

Write three different integer addition expressions that equal –5.

podryga [215]

Answer:

1. 5 + (-10)

2. -8 + 3

3. -7 + 2

5 0

3 years ago

Evaluate the limit of sequence:

mr_godi [17]

Rationalize both the numerator and denominator. Given

$\dfrac{\sqrt a-\sqrt b}{\sqrt c-\sqrt d}$

we can rationalize it by introducing conjugates of the numerator and denominator:

$\dfrac{\sqrt a-\sqrt b}{\sqrt c-\sqrt d} \cdot \dfrac{\sqrt a+\sqrt b}{\sqrt a+\sqrt b} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt c+\sqrt d} \\\\ = \dfrac{\left(\sqrt a\right)^2 - \left(\sqrt b\right)^2}{\left(\sqrt c\right)^2-\left(\sqrt d\right)^2} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt a + \sqrt b} \\\\ = \dfrac{a-b}{c-d} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt a + \sqrt b}$

Then the limit is equivalent to

$\displaystyle \lim_{n\to\infty} \frac{(n+3)-n}{(n+1)-n} \cdot \dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n} = 3 \lim_{n\to\infty} \dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n}$