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Kamila [148]
3 years ago
8

The Cooking Club made some pies to sell during lunch to raise money for an end-of-year banquet. The cafeteria contributed three

pies to the club. Each pie was then cut into seven pieces and sold. There were a total of 91 pieces to sell. How many pies did the club make?
Mathematics
1 answer:
Alex73 [517]3 years ago
3 0
7x13=91 meaning 13 pies were made in total. Since the cafeteria contributed 3 we can subtract 3 from 13. Ultimately meaning that the club contributed 10 pies.

The answer is 10.
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What is the value of (3/10)3
Elenna [48]

Answer:

\frac{9}{10}

Step-by-step explanation:

In this question, one is asked to multiply a fraction by a whole number, first model the question.

\frac{3}{10}(3)

In a situation like this, one multiplies the whole number by the numerator (number) over the fraction.

\frac{3*3}{10}

Simplify,

\frac{9}{10}

A common error that people make is that they will multiply both the numerator and denominator by the whole number, this is the same as multiplying the answer by (1), thus it is incorrect.

6 0
2 years ago
Tom ate 1/2 of the leftovers. How much pie in all did he eat
melisa1 [442]
1/2+1/2=2/2=1 so 1/2 was eaten and the other half was leftovers.
8 0
3 years ago
Read 2 more answers
Solve for x by completing the square: x2 - 12x + 11 = 0
sineoko [7]

Answer:

B

Step-by-step explanation:

x² - 12x + 11 = 0 ( subtract 11 from both sides )

x² - 12x = - 11

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 6)x + 36 = - 11 + 36

(x - 6)² = 25 ( take square root of both sides )

x - 6 = ± \sqrt{25} = ± 5 ( add 6 to both sides )

x = 6 ± 5

Then

x = 6 - 5 = 1 ⇒ (1, 0 )

x = 6 + 5 = 11 ⇒ (11, 0 )

6 0
2 years ago
A group of 100 people is divided into 2 teams with 45 people in team A and 55 people in team B.
Zepler [3.9K]
Team A) 45 people
Team B) 55 people

A)There are two ways to solve this problem, finding the number of combinations possible for Team B, or the number of combinations possible for Team A. 
Team A
It's a given that 20 mathematicians are on team A, which leavs the other 25 people for team A to be chosen from a pool of 80 (100- 20 mathletes)
80-C-25 = 80! / (25!/(80-25)!) =<span>363,413,731,121,503,794,368
</span>or 3.63 x 10^20
Solving using Team B
Same concept, but choosing 55 from a pool of 80 (mathletes excluded)
80-C-25 = 80! / (55!(80-55!) = 363,413,731,121,503,794,368
or 3.63 x 10^20

As you can, we get the same answer for both. 

B)
If none of the mathematicians are on team A, then we exclude the 20 and choose 45:
80-C-45 = 80! / (45!(80-45)!) = <span>5,790,061,984,745,3606,481,440
or 5.79 x 10^22

Note that, if you solve from the perspective of Team B (80-C-35), you get the same answer</span>
7 0
3 years ago
Please Help. Will give brainliest.
Keith_Richards [23]

Answer:

<em>mDBC = 341 °</em>

Step-by-step explanation:

Arc DBC is almost the whole circle, if you were to exclude the degree measure of minor arc DC; which is, in other words m∠ DPC

Now by Vertical Angles Theorem:

m ∠ DPC = m∠ APB, ⇒ provided BD, and AC are diameters

m∠ DPC = 19°

This would mean ⇒

mDBC = 360 - m∠ DPC,

mDBC = 360 - 19,

mDBC = 341 degrees ( ° )

3 0
2 years ago
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