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astraxan [27]
3 years ago
9

Consider the rates 4/5 miles in 8 minutes and 4 minutes to travel 2/5 mile

Mathematics
1 answer:
Anna007 [38]3 years ago
8 0

Answer:

960

Step-by-step explanation:

I'm not sure exactly what's the question, but here's what I found. Lets first put it in our own words.

Consider the speed 4/5 miles in 12 minutes to travel 2/8.

8 + 4 = 12, so you can replace that with 12 minutes.

Now, you can absolutely guess your question.

Time = 12 minutes.

Distance = x (Let x be distance)

Speed = 4/5.

Now, you find distance.

Distance Formula: You can use the equivalent formula d = st which means distance equals rate times time.

So 4/5 × 12, is 9.6.

In fraction, if you have 1 numbers after the decimal point, multiply both numerator and denominator by 10. So 9.6 = (9.6 × 10) × 10 = 960.

Here is a pattern I found

2/5 divided by 4 is 1/10

4/5 divided by 8 is 1/10

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Express sin Q as a fraction in simplest terms.​
Lesechka [4]

Answer:

Solution given

Sin Q=opposite/hypotenuse=28/29

<u>28/29</u><u> </u><u>is</u><u>. </u><u>a</u><u> </u><u>required</u><u> </u><u>answer</u><u>.</u><u />

3 0
3 years ago
If x varies directly with y and x = 3.5 when y = 14, find x when y = 18.
maks197457 [2]

Answer:

x = 4.5

Step-by-step explanation:

there's definitely other ways to solve this, but I used a proportion.

\frac{3.5}{x} = \frac{14}{18}

Then you can make this equation:

14x = 3.5 * 18

I solved for it and got 4.5 .

5 0
3 years ago
At Southtown High school the number of students in band is 1 3/4 times the number in orchestra. If 56 students are in orchestra,
NeTakaya

Answer:

98

Step-by-step explanation:

56 x 1 3/4

56 x 1.75 = 98

8 0
2 years ago
choose the answer that best represents the situation described below liza is buying a new bicycle weighs the more expensive it w
neonofarm [45]

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6 0
3 years ago
Solve the system by substitution.<br> - 8x + y = -32<br> - 3x – 10 = y<br> (,)
agasfer [191]

Answer:

x=2,\:y=-16

Step-by-step explanation:

\begin{bmatrix}-8x+y=-32\\ -3x-10=y\end{bmatrix}\\\\\mathrm{Isolate}\:x\:\mathrm{for}\:-8x+y=-32:\quad x=-\frac{-32-y}{8}\\\\\mathrm{Subsititute\:}x=-\frac{-32-y}{8}\\\begin{bmatrix}-3\left(-\frac{-32-y}{8}\right)-10=y\end{bmatrix}\\\\Simplify\\\begin{bmatrix}\frac{3\left(-32-y\right)}{8}-10=y\end{bmatrix}\\\\\mathrm{Isolate}\:y\:\mathrm{for}\:\frac{3\left(-32-y\right)}{8}-10=y:\quad y=-16\\\\\mathrm{For\:}x=-\frac{-32-y}{8}\\\\\mathrm{Subsititute\:}y=-16\\x=-\frac{-32-\left(-16\right)}{8}\\

-\frac{-32-\left(-16\right)}{8}=2\\x=2\\x=2,\:y=-16

3 0
4 years ago
Read 2 more answers
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