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Olegator [25]
2 years ago
12

Can someone please help me im confsued

Mathematics
1 answer:
sergij07 [2.7K]2 years ago
7 0
Answer: 25,600
explanation: subtracting the bigger number to the lowest number.
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The low temperatures for a week in Edmonton, Alberta are -15 C, -12 C, -10 C, -12 C, -18 C, -20 C, and -25 C. What is the mean l
MakcuM [25]

-15.7

Step-by-step explanation:

2

-15

-12

-10

-12

-18

-20

-25

-------

-110

7 divide by -110=-15.7

7 0
3 years ago
An owner of a key rings manufacturing company found that the profit earned (in thousands of dollars) per day by selling n number
vovikov84 [41]

Answer:

The number of key rings sold on a particular day when the total profit is $5,000 is 4,000 rings.

Step-by-step explanation:

The question is incomplete.

<em>An owner of a key rings manufacturing company found that the profit earned (in thousands of dollars) per day by selling n number of key rings is given by </em>

<em />P=n^2-2n-3<em />

<em>where n is the number of key rings in thousands.</em>

<em>Find the number of key rings sold on a particular day when the total profit is $5,000.</em>

<em />

We have the profit defined by a quadratic function.

We have to calculate n, for which the profit is $5,000.

P=n^2-2n-3=5\\\\n^2-2n-8=0

We have to calculate the roots of the polynomial we use the quadratic equation:

n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\n= \frac{-2\pm\sqrt{4-4*1*(-8)}}{2}= \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{36}}{2} =\frac{-2\pm6}{2} \\\\n_1=(-2-6)/2=-8/2=-4\\\\n_2=(-2+6)/2=4/2=2

n1 is not valid, as the amount of rings sold can not be negative.

Then, the solution is n=4 or 4,000 rings sold.

7 0
3 years ago
I need help on 1-6 and If you can, can you explain how to do these? I'm not entirely sure how to.
gayaneshka [121]
1. Using c=2pi(r), plug in 7 for r and solve. Then using a=pi(r)^2, plug in 7 for r once again and solve.

2. First, the diameter (d) is 12 so to get the radius (r), divide 12 by 2 and you should get 6. Then use c=2pi(r) for circumference and a=pi(r)^2 for area to solve.

3. To get the area of the semicircle, divide 16 by 2 to get the radius (r), plug it into a=pi(r)^2, and divide the answer you get for a by 2. To get the area of the triangle, use a=1/2bh, plugging in 16 for b and 10 for h. Finally, add your two answers (the a's from the semicircle and triangle problems).

4. Multiply 20 by 5.5 to get the area of the triangle. Then multiply 4.5 by 20 to get the area of the parallelogram and add your two quotients.

5. Use a=1/2bh and plug in 4 for b and 3 for h and solve. Then multiply the quotient by 10 and there's your volume. To find the surface area, solve SA=(10×4)+(10×3)+(10×5)+12. All I did there was find the area of all the sides and added them together.

6. To find the triangle's volume, use a=1/2bh (b=4, h=1.5) and then multiply the quotient of that by 2.5. To find the rectangle's volume, use v=lwh (l=4, w=2.5, h=2) and solve. Finally, add the triangle's volume and the rectangle's volume to get the total volume. To get its surface area, start with the rectangle. Find the areas of all the sides and add them together but then subtract the 2.5×4 rectangle as it is not on the surface. It should look like this: SA=2(4×2)+2(2.5×2)+10. Again, all I did was find the areas of all the rectangle's sides on the surface and added them. Next, find the triangle's areas on the surface and it should look like this: SA=2(1.5×4)+2(2.5×2.5). Finally, add both values of SA from the triangle and rectangle and there's your surface area.
7 0
3 years ago
ALGEBRA FINAL HELP PLEASE!!! ASAP!!!
Slav-nsk [51]

Answer: I am pretty sure the answer is

g(x) = (x+3)^2 +9

Consider brainleist?

Hope this helped

:D

8 0
3 years ago
A rectangular swimming pool measures 40 ft by 60 ft and is surrounded by a path of uniform width around the four edges. The peri
Alex_Xolod [135]

Answer:

<em><u>6ft</u></em>

Step-by-step explanation:

<em><u>Lets</u></em><em><u> </u></em><em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>width</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>path</u></em><em><u> </u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>surrounding </u></em><em><u>path</u></em><em><u> </u></em><em><u>wil</u></em><em><u>l</u></em><em><u> </u></em><em><u>add</u></em><em><u> </u></em><em><u>2x</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>pool</u></em><em><u> </u></em><em><u>dimension</u></em><em><u>,therefore</u></em><em><u> </u></em><em><u>over</u></em><em><u> </u></em><em><u>all</u></em><em><u> </u></em><em><u>dimesion</u></em><em><u>:</u></em><em><u> </u></em><em><u>(</u></em><em><u>2x</u></em><em><u>+</u></em><em><u>4</u></em><em><u>0</u></em><em><u>)</u></em><em><u> </u></em><em><u>b</u></em><em><u>y</u></em><em><u> </u></em><em><u>(</u></em><em><u>2x</u></em><em><u>+</u></em><em><u>6</u></em><em><u>0</u></em><em><u>)</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>over</u></em><em><u>all</u></em><em><u> </u></em><em><u>perimeter</u></em><em><u> </u></em><em><u>(</u></em><em><u>2x</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>4</u></em><em><u>0</u></em><em><u>)</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>2</u></em><em><u>(</u></em><em><u>2x</u></em><em><u>+</u></em><em><u>6</u></em><em><u>0</u></em><em><u>)</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>2</u></em><em><u>4</u></em><em><u>8</u></em><em><u> </u></em><em><u>Simplify</u></em><em><u> </u></em><em><u>divide</u></em><em><u> </u></em><em><u>b</u></em><em><u>y</u></em><em><u> </u></em><em><u>2,</u></em><em><u> </u></em><em><u>result</u></em><em><u> </u></em><em><u>(</u></em><em><u>2</u></em><em><u>x</u></em><em><u>+</u></em><em><u>4</u></em><em><u>0</u></em><em><u>)</u></em><em><u> </u></em><em><u>+</u></em><em><u>(</u></em><em><u>2</u></em><em><u>x</u></em><em><u>+</u></em><em><u>6</u></em><em><u>0</u></em><em><u>)</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>2</u></em><em><u>4</u></em>

<em><u> </u></em><em><u>Combine</u></em><em><u> </u></em><em><u>like</u></em><em><u> </u></em><em><u>term</u></em><em><u>s</u></em><em><u> </u></em><em><u>2x</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>2</u></em><em><u>x</u></em><em><u> </u></em><em><u>+</u></em><em><u>4</u></em><em><u>0</u></em><em><u> </u></em><em><u>+</u></em><em><u>6</u></em><em><u>0</u></em><em><u> </u></em><em><u>=</u></em><em><u>1</u></em><em><u>2</u></em><em><u>4</u></em><em><u> </u></em>

<em><u>4x</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>1</u></em><em><u>0</u></em><em><u>0</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>2</u></em><em><u>4</u></em><em><u> </u></em>

<em><u>4x</u></em><em><u>=</u></em><em><u>1</u></em><em><u>2</u></em><em><u>4</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>0</u></em><em><u>0</u></em>

<em><u>4</u></em><em><u>x</u></em><em><u>=</u></em><em><u>2</u></em><em><u>4</u></em>

<em><u>x</u></em><em><u>=</u></em><em><u>2</u></em><em><u>4</u></em><em><u>/</u></em><em><u>4</u></em>

<em><u>x</u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em><em><u>ft</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>th</u></em><em><u>e</u></em><em><u> </u></em><em><u>width</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>path</u></em>

<em><u>check</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>by</u></em><em><u> </u></em><em><u>finding</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>perimeter</u></em><em><u> </u></em><em><u>with</u></em><em><u> </u></em><em><u>these</u></em><em><u> </u></em><em><u>values</u></em><em><u>;</u></em><em><u> </u></em><em><u>2</u></em><em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>12</u></em><em><u> </u></em><em><u>ft</u></em><em><u> </u></em>

<em><u>2</u></em><em><u> </u></em><em><u>(</u></em><em><u> </u></em><em><u>1</u></em><em><u>2</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>4</u></em><em><u>0</u></em><em><u> </u></em><em><u>)</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>2</u></em><em><u>(</u></em><em><u> </u></em><em><u>1</u></em><em><u>2</u></em><em><u> </u></em><em><u>+</u></em><em><u>6</u></em><em><u>0</u></em><em><u> </u></em><em><u>)</u></em><em><u> </u></em>

<em><u>2</u></em><em><u>(</u></em><em><u> </u></em><em><u>5</u></em><em><u>2</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>2</u></em><em><u>(</u></em><em><u>7</u></em><em><u>2</u></em><em><u>)</u></em>

<em><u>1</u></em><em><u>0</u></em><em><u>4</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>1</u></em><em><u>4</u></em><em><u>4</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>2</u></em><em><u>4</u></em><em><u>8</u></em><em><u>;</u></em><em><u> </u></em><em><u>confirms</u></em><em><u> </u></em><em><u>our</u></em><em><u> </u></em><em><u>solution</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>x</u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em><em><u> </u></em><em><u>ft</u></em>

5 0
3 years ago
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