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nikdorinn [45]
3 years ago
11

Write 0.125 as a fraction​

Mathematics
2 answers:
Pani-rosa [81]3 years ago
7 0

Answer:

125/1000= 25/200= 5/40= 1/8

SashulF [63]3 years ago
5 0

Answer:

Hey mate!!

Ur answer is, 0.125= 125/1000

pls mark as BRAINLIST

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There is 1 inch of snow on the ground when it begins to snow 0.75 inch per hour. Which linear equation represents the total dept
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Y=0.75x+1 is the answer
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A cab company charges a flat amount of $5 plus $1.10 per mile traveled. If at the end of a cab ride a customer owes the driver $
Luda [366]

I believe that the number of miles traveled would be 8. This is because you would subtract $5.00 from the $13.80 and then divided by $1.10.


8 0
3 years ago
The diagonal of a rectangle is 25 inch. the width is 15 in. what is the area of the rectangle?
givi [52]
D= diagonal= 25 in
w= 15 in

A= w√(d^2﹣w^2)

A= (15)√(25^2 - 15^2)
square in parentheses

A= (15)√(625 - 225)
subtract in parentheses

A= (15)√(400)
take square root of 400

A= (15)(20)

A= 300 in^2


ANSWER: The area is 300 inches squared

Hope this helps! :)
6 0
3 years ago
Which point is the vertex of the angle below
tigry1 [53]

Answer:

The answer is K

5 0
3 years ago
Read 2 more answers
The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped
kakasveta [241]

Check the picture below.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2  = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0
3 years ago
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