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3 years ago

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If you know the rate of a water leak in gallons per hour, how can you find the number of hours it takes for 1 gallon to leak out

? Justifyyour answer

1 answer:

Evgen [1.6K]3 years ago

7 0

1/the rate of leakage per hour

This will give you the time it takes for 1 gallon to leak out in hours.

For example, if something is leaking at the rate of 12 gallons per hour, it will take 1/12 of an hour for 1 gallon to leak out. ( or 5 min)

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What is the question of the line that passes through the point (-2,-2) and has a slope of 4. Plz someone answer

Studentka2010 [4]

Answer:y=4x+6

Step-by-step explanation:

We have the information that the slope is 4 and the line goes through the point (-2,-2). With this information, we can make a linear equation in a point slope form ( $y - y_{1}= m * (x - x_{1})$ , so the equation would be $y+2=4(x+2)$ , or simplified, $y+2=4x+8.$ in order to solve for y (to make it a slope-intercept equation), we must subtract 2 from both sides. This gives us the equation y=4x+6. Hope this helps!

4 0

3 years ago

Please help its from the lines and angles chapter

Murrr4er [49]

This is the answer of this question

4 0

4 years ago

Rice weighing 3 3/4 pounds was divided equally and placed in 4 containers. How many ounces of rice were in each?

IRINA_888 [86]

Answer:

Solution

To solve the problem first we have simplify the improper fraction.

⇒ 33/4 ÷ 4 pounds.

= (4 × 3 + 3) /4 ÷ 4 pounds.

= 15/4 ÷ 4 pounds.

= 15/4 × 1/4 pounds.

= 15/16 pounds.

Now we know that, 1 pound = 16 ounces.

Therefore, 15/16 pounds .

= 15/16 × 16 ounces.

= 15/1

= 15

= 15 ounces.

Answer: 15 ounces.

4 0

2 years ago

Read 2 more answers

4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513

dimaraw [331]

Answer:

Part 4) $sin(\theta)=\frac{12}{13}$

Part 10) The angle of elevation is $40.36\°$

Part 11) The angle of depression is $78.61\°$

Part 12) $arcsin(0.5)=30\°$ or $arcsin(0.5)=150\°$

Part 13) $-45\°$ or $225\°$

Step-by-step explanation:

Part 4) we have that

$cos(\theta)=-\frac{5}{13}$

The angle theta lies in Quadrant II

so

The sine of angle theta is positive

Remember that

$sin^{2}(\theta)+ cos^{2}(\theta)=1$

substitute the given value

$sin^{2}(\theta)+(-\frac{5}{13})^{2}=1$

$sin^{2}(\theta)+(\frac{25}{169})=1$

$sin^{2}(\theta)=1-(\frac{25}{169})$

$sin^{2}(\theta)=(\frac{144}{169})$

$sin(\theta)=\frac{12}{13}$

Part 10)

Let

$\theta$ ----> angle of elevation

we know that

$tan(\theta)=\frac{85}{100}$ ----> opposite side angle theta divided by adjacent side angle theta

$\theta=arctan(\frac{85}{100})=40.36\°$

Part 11)

Let

$\theta$ ----> angle of depression

we know that

$sin(\theta)=\frac{5,389-2,405}{3,044}$ ----> opposite side angle theta divided by hypotenuse

$sin(\theta)=\frac{2,984}{3,044}$

$\theta=arcsin(\frac{2,984}{3,044})=78.61\°$

Part 12) What is the exact value of arcsin(0.5)?

Remember that

$sin(30\°)=0.5$

therefore

$arcsin(0.5)$ -----> has two solutions

$arcsin(0.5)=30\°$ ----> I Quadrant

or

$arcsin(0.5)=180\°-30\°=150\°$ ----> II Quadrant

Part 13) What is the exact value of $arcsin(-\frac{\sqrt{2}}{2})$

The sine is negative

so

The angle lies in Quadrant III or Quadrant IV

Remember that

$sin(45\°)=\frac{\sqrt{2}}{2}$

therefore

$arcsin(-\frac{\sqrt{2}}{2})$ ----> has two solutions

$arcsin(-\frac{\sqrt{2}}{2})=-45\°$ ----> IV Quadrant

or

$arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\°$ ----> III Quadrant

5 0

3 years ago

Solve the inequality.<br><br> 2 - | 2/5x +3 | > 3/5

meriva

Answer:

Inequality form:

-11< x < -4

Interval Notation:

(-11, -4)

7 0

3 years ago