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suter [353]
2 years ago
13

WHEN CALCULATING THE AREA OF A TRAPEZOID, HOW DO YOU DETERMINE WHICH SIDES ARE THE BASES

Mathematics
1 answer:
dolphi86 [110]2 years ago
6 0

Answer:

a bc they are parallel

Step-by-step explanation:

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which rule would apply to the expression w^2w^3. A.addition rule. B.power rule. C.distributive rule. D.quotient to difference ru
Free_Kalibri [48]
"Power rule" is the one rule among the following choices given in the question that <span>would apply to the expression w^2w^3. The correct option among all the options that are given in the question is the second option or option "B". I hope that this is the answer that has actually come to your desired help.</span>
4 0
3 years ago
Coffee company wants to make sure that their coffee is being served at the right temperature. If it is too hot, the customers co
andrey2020 [161]

Answer:

The average temperature of coffee in the sample, which is 70.2.

Step-by-step explanation:

The population is all the costumers. The company has determined that they want the average coffee temperature to be 65 degrees C. So, 65 degrees C is the average of the population.

The sample is the 20 orders. The mean \mu of the sample is 70.2C(That is, the mean of the 20 orders0. So, the correct answer is:

The average temperature of coffee in the sample, which is 70.2.

3 0
3 years ago
What I would like to know is how to solve this problem
ratelena [41]
30 - 29 = 1 for the first one.
3 0
2 years ago
-140 divided by 5 1/2 then rounded to the nearest 10th of a foot
storchak [24]

Answer:

                                                                                                         

Step-by-step explanation:

8 0
2 years ago
John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
2 years ago
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