12 gallons in 4 minutes ; 21 gallons in 7 minutes

Mathematics

1 answer:

Fofino [41]2 years ago

8 0

1. 3 Gallons per minute

2. 3 gallons per minute

The unit rate is 3 gallons.

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If N/8+6=58, then N equals?

GaryK [48]

N equals <em>416</em>

Hope this helps!

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2 years ago

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4x + 1/3 y 2 what is the value of the expression above when x = 2 and y = 3

GenaCL600 [577]

Hoi!

To solve this, first plug in the values for x and y.

x = 2, so anywhere you see x, put 2 in its place.

y = 3, so anywhere you see y, put 3 in its place.

$4(2) + \frac{1}{2}(3)$

4 × 2 = 8

$\frac{1}{2}$ × 3 = $1 \frac{1}{2}$

$8 +1 \frac{1}{2}$ = $9 \frac{1}{2}$

$9 \frac{1}{2}$ is your answer.

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2 years ago

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two nascar cars are racing around a 2 mile track. one car can complete a lap in 48 seconds, and the other car can complete one l

Citrus2011 [14]

Let's assume that the faster car has a 2 mile "head start".
The faster car is 1.2 times faster (48/40) than the slower car.
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Five 2 mile trips will take 5 * 40 seconds = 200 seconds.

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3 years ago

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Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

omeli [17]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

$=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1$

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

3 0

3 years ago

How to do this ? ....

DerKrebs [107]

I’m pretty sure it’s 11. So answer D

3 0

3 years ago