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Maksim231197 [3]
4 years ago
15

Peter has $4.80 in his piggy bank, consisting of dimes and quarters. If there are 33 coins in all, how many of each does he have

?
Mathematics
1 answer:
yaroslaw [1]4 years ago
6 0

Answer:

10 quarters, 23 dimes

Step-by-step explanation:

Make dimes = d and quarters = q. Let the number of dimes and quarters add up to 33 and the value of the dimes and quarters add up to 4.80. The system of equations should look like:

  • d + q = 33
  • 0.10d + 0.25q = 4.80

Solve for d in the first equation. Subtract q from both sides.

d = 33 - q

Substitute d into the second equation.

0.10(33 - q) + 0.25q = 4.80

Distribute 0.10 inside the parentheses.

3.3 - 0.10q + 0.25q = 4.80

Combine like terms.

3.3 + 0.15q = 4.80

Subtract 3.3 from both sides.

0.15q = 1.5

Divide both sides by 0.15.

q = 10

Substitute this value of q into where we solved for d (d = 33 - q).

d = 33 - (10)

d = 23

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I already have the first question done! I just need help with the other two.
Rainbow [258]

Problem 2

Part 1

I'll replace h with y and replace t with x.

The equation turns into y = -5x^2 + 40x + 45

It's of the form y = ax^2+bx+c where

  • a = -5
  • b = 40
  • c = 45

Use the 'a' and b values to find the value of h, which is the x coordinate of the vertex

h = -b/(2a)

h = -40/(2(-5))

h = -40/(-10)

h = 4

At the four second mark is when the rocket will reach its peak height.

Plug this into the original equation to find its paired y value

y = -5x^2 + 40x + 45

y = -5(4)^2 + 40(4) + 45

y = 125

The vertex is at (h,k) = (4, 125).

<h3>The highest the rocket goes is 125 feet.</h3>

---------------

Part 2

Plug in y = 0 and solve for x to find when the rocket hits the ground. I'll use the quadratic formula.

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(40)\pm\sqrt{(40)^2-4(-5)(45)}}{2(-5)}\\\\x = \frac{-40\pm\sqrt{2500}}{-10}\\\\x = \frac{-40\pm50}{-10}\\\\x = \frac{-40+50}{-10} \ \text{ or } \ x = \frac{-40-50}{-10}\\\\x = \frac{10}{-10} \ \text{ or } \ x = \frac{-90}{-10}\\\\x = -1 \ \text{ or } \ x = 9\\\\

Ignore the negative x value. We cannot have negative time values.

The only practical root is that x = 9, meaning the rocket hits the ground at the 9 second mark.

<h3>The rocket is in the air for 9 seconds.</h3>

=====================================================

Problem 3

You have a product that sells for $10 and 1000 people buy per month.

The revenue you pull in based on those figures is 10*1000 = 10,000 dollars per month.

If you raised the price by $1, then the $10 jumps to $11. The downside is that the 1000 people drops to 900 people (you lose 100 customers). The revenue would be 11*900 = 9900. So far, it seems like a bad idea to raise prices. But we'll increase the price once more to see what happens.

If the price goes to $12, then you lose another 100 customers and you now have 800 customers. So that's 12*800 = 9600 dollars in revenue per month. It seems the trend is getting worse.

---------------

Let's generalize what's going on.

x = number of times you raise the price by $1

The old price is $10 per item. It jumps up to 10+x per item. At the same time, the customer count goes from 1000 to 1000-100x. Each time x goes up by 1, the expression 1000-100x goes down by 100.

Multiply the price and customer count to get the amount earned.

revenue = (price)*(number of customers)

revenue = (10+x)*(1000-100x)

revenue = 10(1000 - 100x) + x(1000-100x)

revenue = 10,000 - 1000x + 1000x - 100x^2

revenue = 10,000 - 100x^2

revenue = -100x^2 + 10,000

If you were to use the methods done in the previous problem, you should find the vertex is at (0,10000). This means that the max revenue was already reached when x = 0 price increases were done. This reinforces the previous results we got earlier before we started generalizing in terms of x.

You should <u>not</u> raise the price, or else you'll just continue to lose customers until you go to 0. No amount of price raising will get your revenue up, which in turn means the profits will suffer as well. With the product at $10, you are already at the max revenue point.

<h3>Conclusion: Keep the price at $10</h3>
3 0
3 years ago
What’s the answer to this?
Rina8888 [55]

Answer:

The answer in order from least to greatest is C.

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=a%20-%207%20%3D%2017" id="TexFormula1" title="a - 7 = 17" alt="a - 7 = 17" align="absmiddle" c
andriy [413]
A would equal 24 because 17+7 is 24
6 0
3 years ago
A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must h
mestny [16]

Answer:

Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed  = Volume of the two Hemisphere + Volume of the Cylinder

Volume of a Hemisphere  =\frac{2}{3}\pi r^3

Volume of a Cylinder =\pi r^2 h

Therefore:

The Volume of the Solid formed

=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Area of the Hemisphere =2\pi r^2

Curved Surface Area of the Cylinder =2\pi rh

Total Surface Area=

2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh

Cost of the Hemispherical Ends  = 2 X  Cost of the surface area of the sides.

Therefore total Cost, C

=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh

Recall: h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Therefore:

C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}

The minimum cost occurs at the point where the derivative equals zero.

C^{'}=\frac{-27840+32\pi r^3}{3r^2}

When \:C^{'}=0

-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518

Recall:

h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet

Therefore, the dimensions that will minimize the cost are:

Radius =6.518 feet

Height = 26.074 feet

5 0
3 years ago
-3х + Зу = 6<br><br> A.no solution<br> B.one solution<br> C.two solutions<br> D.infinite solutions
7nadin3 [17]
A it’s no solutions hope I helped!
5 0
3 years ago
Read 2 more answers
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