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kirill115 [55]
3 years ago
7

the volleyball team needs new uniforms. the students plan to sell plush toy Eagles ( the school mascot) for $5 each. The student

s find three companions on-line that sell stuff mascots. Company A cell's 12 eagles for $34.08. Company B cells 15 equals for $42.90. Company C charges $50.58 for 18 eagles. which company has the Best Buy?​
Mathematics
2 answers:
In-s [12.5K]3 years ago
8 0
The correct answer would be c
sammy [17]3 years ago
7 0

Answer:

Company C

Step-by-step explanation:

You divide the amount of money it all costs to the number of plushies they are selling it for.

A: 34.08/12 = $2.84 each

B: 42.90/15 = $2.86 each

C: 50.58/18 = $2.81 each

So C has the least amount of money spent for a greater amount of plushies so C would be the Best Buy.

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Given the following definitions:
sveticcg [70]

Step-by-step explanation:

First,

(A intersect B) = {1,2,4,5} intersect {1,3,5,7}

= {1,5}

Now,

A'n B = (A) - (A intersect B)

= {1,2,4,5} - {1,5}

= {2,4}

7 0
3 years ago
One light-year equals 5.9 x 1012 miles. How many light-years are in 6.79
inna [77]

Option B

The number of light years in 6.79 \times 10^{16} miles is 11508 light years

<em><u>Solution:</u></em>

Given that,

One light-year equals 5.9 x 10^12 miles

Therefore,

1 \text{ light year } = 5.9 \times 10^{12} \text{ miles }

To find: Number of light years in 6.79 \times 10^{16} miles

Let "x" be the number of light years in 6.79 \times 10^{16} miles

Then number of light years in 6.79 \times 10^{16} miles can be found by dividing 6.79 \times 10^{16} miles by miles in 1 light year

\text{Number of light years in } 6.79 \times 10^{16} miles = \frac{6.79 \times 10^{16}}{5.9 \times 10^{12}}\\\\\text{Use the law of exponent }\\\\\frac{a^m}{a^n} = a^{m-n}\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = \frac{6.79}{5.9} \times 10^{16-12}\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = 1.1508 \times 10^4\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = 11508

Thus number of light years in 6.79 \times 10^{16} miles is 11508 light years

3 0
3 years ago
Please help me answer these 2 questions, worth lots of points
alexgriva [62]
For the first one, to find x, you are going to take that entire side which would equal to 180 and solve. 2x+20+55=180 would then go down to 2x+65=180 once you add the 20 and 55. Then subtract 65 to both sides and divide your final answer 115/2 which is x=57.5. For the second one it’s 4x-2+21=180, then you will subtract 2 from 21 and get 19. After you would subtract 19 by 180 and divide 4 by each side, getting x=40.25 as your final answer.
7 0
3 years ago
Will mark Brainliest!! :)
Georgia [21]

Step-by-step explanation:

The system of equations for eq 1 which is 3x + y = 118 represents the Green High School which filled three buses(with a specific number of students identified as x) and a van(with a specific number of students identified as y) with a total of 118 students.

for eq 2; 4x + 2y = 164; represents Belle High School which filled four buses(with a specific number of students identified as x) and two vans(with a specific number of students identified as y) with a total of 164 students.

The solution represents the specific number of students in the buses and vans in eq1 and eq 2 with x being 36 students and y being 10 students.

substituting 36 for x and 10 for y in eq 1;

3(36) + 10 = 108 + 10 = 118 total students for Green High School

substituting 36 for x and 10 for y in eq2;

4(36) + 2(10) = 144 + 20 = 164 total students for Belle High school

6 0
2 years ago
A manufacturer produces bolts of a fabric with a fixed width. A quantity q of this fabric (measured in yards) that is sold is a
BigorU [14]

Answer:

R'(20) = 2000

Step-by-step explanation:

We are given the following in the question:

Quantity, q

Selling price in dollars per yard, p

q=f(p)

Total revenue earned =

R(p)=pf(p)

f(20)=13000

This means that 13000 yards of fabric is sold when the selling price is 20 dollars per yard.

f′(20)=−550

This means that increasing the selling price by 1 dollar per yards there is a decrease in fabric sales by 550.

We have to find R'(20)

Differentiating the above expression, we have,

R'(p) = \displaystyle\frac{d(R(p))}{dp} = \frac{d(pf(p))}{dp} = f(p) + pf'(p)

Putting the values, we get,

R'(p) = f(p) + pf'(p)\\\\R'(20) = f(20) + 20(f'(20))\\\\R'(20) = 13000 + 20(-550) = 2000

7 0
3 years ago
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