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Verizon [17]
3 years ago
15

A pipe s leaking at 1.5 cups per day how many gallons per week is the pipe leaking

Mathematics
1 answer:
kkurt [141]3 years ago
8 0
1.77 so i wish u good lock
You might be interested in
0.79 x 3.7 =<br> please explain your process
mart [117]

This question is super easy.

0.79*3.7

=2.923

Hope this helps!

Thanks!

-Charlie

3 0
3 years ago
A car salesman receives a base weekly salary of $150 plus a commission of $325 per car sold. What is the equation that models th
Temka [501]

Given :

Weekly salary , S = $150.

Commission per car , C = $325.

To Find :

The the equation that models the weekly salary of a car salesman.

Solution :

We know it is a quadratic equation :

Let , the equation is , y = mx + c

Here, x is number of car sold in a week.

For  x = 0 ( He will not get commission)

So , Salary is $150 = c .

For x = 1

Salary is $( 150 + 325 ) = $475 = m(1) + c

Therefore , m = $325 and c = $150.

Required linear equation is y = 325x + 150.

Hence, this is the required solution.

5 0
3 years ago
If 43 of the 148 reams of paper purchased by a department are used, what is the percentage that remains?
Wewaii [24]
100%/x%=148/43
<span>(100/x)*x=(148/43)*x       - </span>we multiply both sides of the equation by x
<span>100=3.44186046512*x       - </span>we divide both sides of the equation by (3.44186046512) to get x
<span>100/3.44186046512=x </span>
<span>29.0540540541=x </span>
<span>x=29.0540540541

</span>now we have: 
<span>43 is 29.0540540541% of 148</span>
3 0
3 years ago
Read 2 more answers
Determine whether each given value of x satisfies the inequality x+1&lt;x/3
Vinil7 [7]

x+1 < \dfrac{x}{3}\ \ \ \ |\cdot3\\\\3x+3 < x\ \ \ \ |-3\\\\3x < x-3\ \ \ \ |-x\\\\2x < -3\ \ \ \ |:2\\\\x < -1.5

Answer: (c) x = -4

5 0
3 years ago
Find the derivative of 3e^y+x=y
Sindrei [870]

You have the following function:

y=3e^y+x

Derivate implictly the previous expression, as follow:

y^{\prime}=3e^yy^{\prime}+1

Where you have used that:

(e^y)^{\prime}=e^yy^{\prime}

Then, the implicit derivative of the given expression is:

y^{\prime}=3e^yy^{\prime}+1

Next, solve for y' as follow:

\begin{gathered} y^{\prime}-3e^yy^{\prime}=1 \\ (1-3e^y)y^{\prime}=1 \\ y^{\prime}=\frac{1}{1-3e^y} \end{gathered}

4 0
1 year ago
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