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stich3 [128]
3 years ago
11

One baseball team won 15 games throughout their entire season. Of all their games, this

Mathematics
1 answer:
Alex_Xolod [135]3 years ago
3 0

Answer:

20

Step-by-step explanation:

To get the answer, you have to find what number 15 is 75 percent of.

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If
Rudiy27
Looking at this problem in terms of geometry makes it easier than trying to think of it algebraically.

If you want the largest possible x+y, it's equivalent to finding a rectangle with width x and length y that has the largest perimeter.

If you want the smallest possible x+y, it's equivalent to finding the rectangle with the smallest perimeter.

However, the area x*y must be constant and = 100.

We know that a square has the smallest perimeter to area ratio. This means that the smallest perimeter rectangle with area 100 is a square with side length 10. For this square, x+y = 20.

We also know that the further the rectangle stretches, the larger its perimeter to area ratio becomes. This means that a rectangle with side lengths 100 and 1 with an area of 100 has the largest perimeter. For this rectangle, x+y = 101.

So, the difference between the max and min values of x+y = 101 - 20 = 81.
6 0
3 years ago
Can you guys help me ?
AysviL [449]

Answer:

C the 12th root of (8^x)

Step-by-step explanation:

This becomes 8 ^ x/4  ^ 1/3

We know that a^b^c = a^ (c*c)

8 ^(x/12)

8 0
3 years ago
Read 2 more answers
Help plzzzzzzzzzzzzzzzz
agasfer [191]

Answer:

One pound of raw chicken will serve approximately four people. or 12 ounces

so A

Step-by-step explanation:

5 0
2 years ago
Alvin is heating a solution in his chemistry lab. The temperature of the solution starts at 14°C. He turns on the bumer and find
iren2701 [21]

Answer:

C. (3,20)

Step-by-step explanation:

5 0
3 years ago
Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15.
Naya [18.7K]
It's a geometric sequence.

4,-12,36,... \\ \\
a_1=4 \\
r=\frac{a_2}{a_1}=\frac{-12}{4}=-3 \\ \\
a_n=a_1 \times r^{n-1} \\
a_n=4 \times (-3)^{n-1} \\
a_n=4 \times (-3)^{-1} \times (-3)^n \\
a_n=4 \times (-\frac{1}{3}) \times (-3)^n \\
a_n=-\frac{4}{3}(-3)^n

It's the sum for term 4 through term 15.

 \boxed{ \sum\limits_{n=4}^{15} (\frac{4}{3}(-3)^n)}
8 0
2 years ago
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