All categories

3 years ago

A water sample must be taken from water at least 20 feet deep. Find the depth of the water 50 feet from shore.

Mathematics

1 answer:

rewona [7]3 years ago

5 0

Answer:

The answer is 15 ft.

Step-by-step explanation:

I hope this helps

You might be interested in

Alex lady sketch for a homemade soccer goal he plans to build . Do you go will be in the shape of a triangular prism . The legs

Otrada [13]

The triangular prism has a volume of
.. V = Bh = (1/2)*(4ft *8 ft)*(24 ft) = 384 ft^3

7 0

3 years ago

Determine the mean, median, and more for the given data in each graph.

77julia77 [94]

5 is you median
4 is your mean

6 0

3 years ago

51 is 33 1/3 of what number?

nevsk [136]

Answer:

$\boxed{\frac{153}{100}\:\:Or\:\:1.53}$

Step-by-step explanation:

Let the number be $n$ .

Then,

$33\frac{1}{3} \tmes n=51$

This implies that;

$\frac{100}{3} n=51$

We multiply both sides by $\frac{3}{100}$ to obtain;

$\frac{3}{100} \times \frac{100}{3}n=\frac{3}{100} \times 51$

$\Rightarrow n=\frac{3}{100} \times 51$

$\Rightarrow n=\frac{153}{100}$

$\Rightarrow n=1.53$

5 0

3 years ago

Read 2 more answers

Katrina drinks 0.5 gallons of water per day. Which expression shows how to find the number of cups of water she drinks in a week

Vilka [71]

The answer is 128 i hope that helps

8 0

3 years ago

Read 2 more answers

To two decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x equals the quantity 3x

Fiesta28 [93]

Given function is

$f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.$

now we need to find the value of k such that function f(x) continuous everywhere.

We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.

So we just need to find both left and right hand limits then set equal to each other to find the value of k

To find the left hand limit (LHD) we plug x=-4 into 3x+k

so LHD= 3(-4)+k

To find the Right hand limit (RHD) we plug x=-4 into

$kx^2-5$

so RHD= $k(-4)^2-5$

Now set both equal

$k(-4)^2-5=3(-4)+k$

$16k-5=-12+k$

$16k-k=-12+5$

$15k=-7$

$k=-\frac{7}{15}$

k=-0.47

<u>Hence final answer is -0.47.</u>

7 0

3 years ago