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Yakvenalex [24]
3 years ago
6

Explain how you can use multiplication to show that two-fourths and four-eighths are equivalent.

Mathematics
1 answer:
valkas [14]3 years ago
8 0

Two-fourths and four-eights are equivalent since they are integer multiples of each other.

<h2 /><h2>Equivalent fractions</h2>

Equivalent fractions are fractions whose values are equal and are integral multiples of each other. That is, one fraction is gotten from the other by either multiplying or dividing its numerator and denominator by the same integer.

For example, 2/5 = 4/10

Since 2/5 = 2 × 2/(5 × 2) = 4/10

Also, 4/10 = 4 ÷ 2/(10 ÷ 2) = 2/5

So, both 2/5 and 4/10 are equivalent fractions since we either multiply or divide both the numerator and denominator of one to get the other.

<h3 /><h3>To show that two-fourths and four-eights are equivalent,</h3>

So, to show that two-fourths and four-eights are equivalent,

two-fourths = 2/4 and four-eights = 4/8.

To show that these are equivalent, we need to show that their numerator and denominators are integer multiples of each other.

So, 2/4 = 2 × 2/(4 × 2) = 4/8

Also, 4/8 = 4 ÷ 2/(8 ÷ 2) = 2/4

Since 2 × 2/(4 × 2) = 4/8, we multiply 2/4 by 2/2 to get 4/8.

So, two-fourths and four-eights are equivalent since they are integer multiples of each other.

Learn more about equivalent fractions here:

brainly.com/question/17220365

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5 0
4 years ago
Find the integral, using techniques from this or the previous chapter.<br> ∫x(8-x)3/2 dx
Soloha48 [4]

Answer:

\int x(8-x)^{3/2}dx= -\frac{16}{5} (8-x)^{\frac{5}{2}} +\frac{2}{7} (8-x)^{\frac{7}{2}} +C

Step-by-step explanation:

For this case we need to find the following integral:

\int x(8-x)^{3/2}dx

And for this case we can use the substitution u = 8-x from here we see that du = -dx, and if we solve for x we got x = 8-u, so then we can rewrite the integral like this:

\int x(8-x)^{3/2}dx= \int (8-u) u^{3/2} (-du)

And if we distribute the exponents we have this:

\int x(8-x)^{3/2}dx= - \int 8 u^{3/2} + \int u^{5/2} du

Now we can do the integrals one by one:

\int x(8-x)^{3/2}dx= -8 \frac{u^{5/2}}{\frac{5}{2}} + \frac{u^{7/2}}{\frac{7}{2}} +C

And reordering the terms we have"

\int x(8-x)^{3/2}dx= -\frac{16}{5} u^{\frac{5}{2}} +\frac{2}{7} u^{\frac{7}{2}} +C

And rewriting in terms of x we got:

\int x(8-x)^{3/2}dx= -\frac{16}{5} (8-x)^{\frac{5}{2}} +\frac{2}{7} (8-x)^{\frac{7}{2}} +C

And that would be our final answer.

8 0
3 years ago
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GrogVix [38]

1 11/25

= 36/25 = (6/5)^2

and  

3 3/8

= 27/8 = (3/2)^3

5 0
3 years ago
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IRINA_888 [86]

Answer:

2

Step-by-step explanation:

It is the only one that makes sense

7 0
3 years ago
Read 2 more answers
If G is the midpoint of FH find FG<br>fg=11x-7 gh= 3x+9
gavmur [86]

Solution:

we are given that

If G is the midpoint of FH,  it mean that

FG=GH

we are also given that

FG=11x-7 ,GH= 3x+9

So we can write

11x-7=3x+9\\&#10;\\&#10;11x-3x=7+9\\&#10;\\&#10;8x=16\\&#10;\\&#10;x=2

So FG=11x-7=11*\frac{1}{2}-7=11*2-7=15


3 0
4 years ago
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