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IrinaK [193]
3 years ago
15

You are planning a fundraiser for your student council. The fundraiser is a Glow in the Dark Dance.

Mathematics
1 answer:
Ganezh [65]3 years ago
8 0

Answer: p x 7

Step-by-step explanation:

Number of people times 7 (the amount of money paid)

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Miss Garry played the Game of Life during Spring Break and kept track of the results from all the times she spun the wheel. Duri
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The probability in faction form would be 25/80 simplified would be 5/16 chance of spinning a 3. The decimal form would be a 0.3125 chance in spinning a 3
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Allison was left an inheritance from her great aunt on her 12 birthday. However, she , she not able to touch it until she is 21
olya-2409 [2.1K]

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18% intrest

Step-by-step explanation:

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3 years ago
2 pairs of numbers whose Least common multiple is the product
IgorC [24]
I'm not sure what you're really asking but 6 and 9's LCM is 18
7 0
3 years ago
B. MRS + mST = MRT Q R. S​
babunello [35]

Answer:

MRS is the demand side of equation while MRT is for the supply side.

MRS defines how much a consumer is willing to give up of good X for 1 additional unit of good Y to stay on the same utility level. It is shown by indifference curve. MRS = Price of X/ Price of Y

Similarly, MRT is how much a supplier is willing to give up producing good X for 1 additional unit of good Y. It is shown by Production Possibility Frontier. MRT = MC of X/ MC of Y

8 0
2 years ago
Suppose that a function​ f(x) is defined for all real values of x except at xequals=c. can anything be said about the existence
Margaret [11]

we are given that

f(x) is defined for all values of x except at x=c

Limit may or may not exist

case-1:

If there is hole at x=c , then limit exist

case-2:

If there is vertical asymptote at x=c , then limit does not exist

Examples:

case-1:

\lim_{x \to c} \frac{x^2-cx}{(x-c)}

We can simplify it

\lim_{x \to c} \frac{x(x-c)}{(x-c)}

=\lim_{x \to c} x

=c

so, we can see that limit exist and it's value defined

case-2:

\lim_{x \to c} \frac{1}{(x-c)}

Left limit is

\lim_{x \to c-} \frac{1}{(x-c)}

=-\infty

Right Limit is

\lim_{x \to c+} \frac{1}{(x-c)}

=+\infty

so, we can see that left limit is not equal to right limit

so, limit does not exist

4 0
3 years ago
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