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uysha [10]
2 years ago
14

What’s 112/250 simplified

Mathematics
2 answers:
Nutka1998 [239]2 years ago
5 0

Answer:

The simplest form of 112/250 is 56/125. Steps to simplifying fractions. Find the GCD (or HCF) of numerator and denominator. GCD of 112 and 250 is 2

svp [43]2 years ago
3 0

Answer:

\frac{56}{125}

Step-by-step explanation:

112/250

Reduce the fraction \frac{112}{250} to the lowest terms by extracting and canceling out 2.

\frac{56}{125}

Hope it helps and have a great day! =D

~sunshine~

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Which statements are correct interpretations of this graph?
kupik [55]

Answer:

B. 3 pages are edited every five minutes

D. 6/10 of a page is edited per minute

Step by step:

Three pages are done at an interval of five minutes.

Six tenths of a page is done every minute

0.6 * 5 = 3 per five minutes

The other statements are false.

12/3 = 4, four done every minute, really?

5 pages are edited every three minutes.

This would disprove statement B.

And does not align with the graph.

Hope this helps.

7 0
3 years ago
Wite a numerical expression for the below problems. Simplify the expressions.
ziro4ka [17]

Answer:

1) 9

2) 34

Step-by-step explanation:

For the first one, it would be

= (6×3)/2

= 18/2

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The second one is,

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4 0
2 years ago
What is the slope of the line graphed on the coordinate plane
Mice21 [21]

Answer:

The slope = 3/4

Step-by-step explanation:

m=\frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

m=\frac{6-3}{2-(-2)}=\frac{6-3}{2+2}

m = 3/4

5 0
2 years ago
Read 2 more answers
Which equation would best help solve the following problem?
timurjin [86]
Y = yo + Vot - (gt^2)/2

0 = 0 + 31t - 4.9 t^2

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Answer: option A.
6 0
3 years ago
Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE
EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

g(y) = \frac{y-5}{y^2-3y+15}

According to the quotient rule:

d(\frac{f(y)}{h(y)})  = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}

Applying the quotient rule:

g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}

The values for which g'(y) are zero are the critical points:

g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10

The critical values are y = 0 and y = 10.

5 0
3 years ago
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