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3 years ago

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A river is rising at a rate of 3 inches per hour. If the river rises more than 2 feet, it will exceed flood stage. How long can

the river rise at this rate without exceeding flood stage?

2 answers:

Butoxors [25]3 years ago

8 0

Answer: 8 hours.

Step-by-step explanation:

Given:- A river is rising at a rate of 3 inches per hour.

⇒ In one hour, the increase in the level of river= 3 inches

If the river rises more than 2 feet, it will exceed flood stage.

⇒ the level of river must be less than 2 feet

We know that, 2 feet=2×12 inches [1 foot= 12 inches]

⇒ 2 feet = 24 inches

Thus time taken to rise at the same rate without exceeding flood stage

= i.e. Time taken to reach 24 inches= $\frac{24}{3}$

= 8 hours.

inn [45]3 years ago

7 0

There are 24 inches in 2 feet. Every hour it rises 3 inches. So 24/3=8. It can rise for 8 hours before exceeding flood stage.

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Read 2 more answers

Which of the points is a solution to the following system of equations? -5x- 3/2y=15 3x+5/6 y=-44/3

tatiyna

<em>Note: Your question sounds a little unclear, but I am assuming that your system of equations is:</em>

$-5x-\:\frac{3}{2}y=15$

$3x+\frac{5}{6}y=-\frac{44}{3}$

It would anyways clear your concept because the procedure to find the solutions remains the same for any set of a system of equations.

Answer:

The solution of the system of equations be:

$x=-\frac{57}{2},\:y=85$

Step-by-step explanation:

Given the system of equations

$-5x-\:\frac{3}{2}y=15$

$3x+\frac{5}{6}y=-\frac{44}{3}$

solving the system of equations

$\begin{bmatrix}-5x-\frac{3}{2}y=15\\ 3x+\frac{5}{6}y=-\frac{44}{3}\end{bmatrix}$

$\mathrm{Multiply\:}-5x-\frac{3}{2}y=15\mathrm{\:by\:}3\:\mathrm{:}\:\quad \:-15x-\frac{9}{2}y=45$

$\mathrm{Multiply\:}3x+\frac{5}{6}y=-\frac{44}{3}\mathrm{\:by\:}5\:\mathrm{:}\:\quad \:15x+\frac{25}{6}y=-\frac{220}{3}$

so the system of equations becomes

$\begin{bmatrix}-15x-\frac{9}{2}y=45\\ 15x+\frac{25}{6}y=-\frac{220}{3}\end{bmatrix}$

adding the equations

$15x+\frac{25}{6}y=-\frac{220}{3}$

$+$

$\underline{-15x-\frac{9}{2}y=45}$

$-\frac{1}{3}y=-\frac{85}{3}$

so

$\begin{bmatrix}-15x-\frac{9}{2}y=45\\ -\frac{1}{3}y=-\frac{85}{3}\end{bmatrix}$

solving $-\frac{1}{3}y=-\frac{85}{3}$ for y

$-\frac{1}{3}y=-\frac{85}{3}$

Multiply both sides by -3

$\left(-\frac{1}{3}y\right)\left(-3\right)=\left(-\frac{85}{3}\right)\left(-3\right)$

$y=85$

$\mathrm{For\:}-15x-\frac{9}{2}y=45\mathrm{\:plug\:in\:}y=85$

$-15x-\frac{9}{2}\cdot \:85=45$

$\mathrm{Add\:}\frac{765}{2}\mathrm{\:to\:both\:sides}$

$-15x-\frac{765}{2}+\frac{765}{2}=45+\frac{765}{2}$

$-15x=\frac{855}{2}$

$\mathrm{Divide\:both\:sides\:by\:}-15$

$\frac{-15x}{-15}=\frac{\frac{855}{2}}{-15}$

$x=-\frac{57}{2}$

Therefore, the solution of the system of equations be:

$x=-\frac{57}{2},\:y=85$

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3 years ago