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Vera_Pavlovna [14]
3 years ago
14

1 3/8 ÷ 1 2/3 quotient in lowest terms

Mathematics
2 answers:
Ivahew [28]3 years ago
3 0

Answer:

=33/40

Step-by-step explanation:

Convert any mixed numbers to fractions.

Then your initial equation becomes:

118÷53

Applying the fractions formula for division,

=11×38×5

=33/40

=33/40

Alisiya [41]3 years ago
3 0
I have the same question and I like ur user
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If the length of AC equals 50, what is the length of the midsegment DE?
Helga [31]

Answer:

Length of DE is 25.

Step-by-step explanation:

Right, so this is something called midsegment theorem. ΔDBE is similar (but not congruent) to ΔABC because of the angles they share. Since DE is described as a midsegment, this means BE=EC and BD=DA; so ΔDBE is half the size of ΔABC, meaning its side lengths are also half that of the larger triangle.

(Sorry if this information was unnecessary. Just got an answer deleted because I wasn't detailed enough.)

5 0
3 years ago
60% of what number is 24
ozzi
Your answer is 14.4
6 0
3 years ago
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Christina has 50 coins consisting of quarters and dimes. The coins combined value is $9.50. Write a system of equation to determ
Doss [256]

Answer:

Christina has 30 quarters and 20 dimes.

Step-by-step explanation:

Given that a quarter is worth 0.25 dollars and a dime is worth 0.10 dollars, if Christina has 50 coins and these add up to a total of $ 9.50, to determine the number of quarters and dimes that she has in her possession, the following calculation must be carried out, taking into account that the difference between a quarter and a dime is 0.15 dollars:

50 x 0.25 = 12.5

12.5 - 9.5 = 3

Thus, the difference between the value that Christina has and the value that she would have if all of her coins were quarters is 3 dollars. Thus, since 0.15 x 20 equals 3, of the 50 coins Christina has, 30 are quarters and 20 are dimes.

This is confirmed through the following equation:

30 x 0.25 + 20 x 0.1 = X

7.5 + 2 = X

9.5 = X

4 0
3 years ago
80% out of 1000 is what
exis [7]
80% of 1000 is 800 hope this helps :)
5 0
4 years ago
Read 2 more answers
One morning, John drive 5 hours before stopping to eat. After lunch, he. increased his speed by 10 mph. If he completed a 500-mi
Marrrta [24]
So  he drove 5hrs, before lunch, now, he was going at speed, say "r", after he ate, he sped up by 10mph, so, his rate is whatever "r" is, plus 10, or " r + 10 ", after lunch

we know the whole trip took 10hrs, so, is 5hrs before lunch and 5hrs after lunch

we also know the whole trip was 500miles, so if he drove, say "d" miles before lunch, after lunch he drove the slack after lunch, or " 200 - d "

recall, your d = rt, distance = rate * time

\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{before lunch}&d&r&5\\
\textit{after lunch}&500-d&r+10&5
\end{array}
\\\\\\

\begin{cases}
\boxed{d}=5r\\
500-d=(r+10)5\\
----------\\
500-\boxed{5r}=(r+10)5
\end{cases}
\\\\\\
\cfrac{500-5r}{5}=r+10\implies 100-r=r+10

and I'm sure you know what that is
6 0
3 years ago
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