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MArishka [77]
3 years ago
11

At a sporting event concession stand, a small bottle of water costs $2.50 and a large bottle costs $3.00. The concession stand e

mployees collected a total of $357 from water bottle sales. They sold twice as many large bottles as small bottles. Let s represent the number of small bottles sold and let l represent the number of large bottles sold. How many of each size water bottle did they sell? (please explain full work)
Mathematics
1 answer:
timama [110]3 years ago
8 0
The answer will be 7.50 because you are doing your times if you do divid its going to be hard so times to get your answer.
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Find the area of this circle use 3.14
sashaice [31]

Answer:

What circle? Can you add a picture of it?

Step-by-step explanation:

5 0
3 years ago
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What is the value of 1/16 times 16 plus negative 16 plus 16?
guapka [62]

Absolute value (no negatives)

1/16=.0625*16=1+16=17+16=33

7 0
3 years ago
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Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th
jonny [76]

Answer:

The answer is \sqrt{\frac{6}{5}}

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy

Solving:

\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy

[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.

Replacing the limits:

6*\frac{1}{3} =2

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ \frac{1}{m}

Solving the double integral with these new limits we have:

\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy

This part is a little bit tricky so let's solve the integral first for dy:

\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx

Replacing the limits:

\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx

Solving now for dx:

[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.

Replacing the limits:

\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}

As I mentioned before, this volume is equal to 1, hence:

\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }

3 0
3 years ago
Here's the deal. I give you a question, you give me an answer. I give you points and Brainliest if possible. You use the points
olga55 [171]

Answer:

f(x) = -(x-3)^2 + 4

Step-by-step explanation:

(h; k) are the coordinates of the vertex.

On the graph are (3, 4), therefore we have:

f(x) = a(x-3)^2 + 4

We have the x- intercept (1,0) -> x=1; y=0.

Substitute them into the equation:

0 = a(1 - 3)^2 + 4

0 = a(-2)^2 + 4

4a + 4 = 0 | -4

4a = -4   |-4

a = -1

So, we have the answer:

f(x) = -(x-3)^2 + 4

5 0
3 years ago
Need more help 5/6 divided by 2
podryga [215]

Answer:

5/12

Step-by-step explanation:

5/6 divided by 2

= 5/6 divided by 2/1

= 5/6 x 1/2

= 5 x 1 over 6 x 2

= 5 over 6 x 2

= 5 / 12

8 0
2 years ago
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