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frutty [35]
3 years ago
14

RS has a midpoint at M(7, 10). Point R is at (8, 10). Find the coordinates of point s.

Mathematics
1 answer:
Elena L [17]3 years ago
5 0

Answer:

S(6, 10)

Step-by-step explanation:

1. The Midpoint (in this case, M) will always be halfway between both R and S (or other characters in some cases).

2. As M is only one X value away from R, it is only 1 X value from S as well, but in the other direction.

3. (7, 10) - (1, 0) = (6, 10)

The (7, 10) is the Midpoint Coordinates.

The (1, 0) is the distance from M to R.

The (6, 10) is the coordinates of S.

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Answer:

Well first of all, one of the numbers is bigger than the other (has greater value).

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6 0
3 years ago
Alpha printing will make 400 brochures for $100. Omega printing will make 1,000 brochures for $100. How much money will you save
Dmitriy789 [7]

it is given in the question that

Alpha printing will make 400 brochures for $100. Omega printing will make 1,000 brochures for $100.

For Alpha, cost of 1 brochure is

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3 years ago
A leprechaun places a magic penny under a girls pillow. The next night there are 2 magic pennies under her pillow. The following
kiruha [24]

Answer:

<em>After </em><em>47</em><em> days she will have more than 90 trillion pennies.</em>

Step-by-step explanation:

At the beginning there was 1 penny. At the second day the amount of pennies under the pillow became 2.

The amount of pennies doubled each day. So the series is,

1,2,4,8,16,32,.....

This series is in geometric progression.

As the pennies from each of the previous days are not being stored away until more pennies magically appear so the sum of series will be,

S_n=\dfrac{a(r^n-1)}{r-1}

where,

a = initial term = 1,

r = common ratio = 2,

As we have find the number of days that would elapse before she has a total of more than 90 trillion, so

\Rightarrow 90\times 10^{12}\le \dfrac{1(2^n-1)}{2-1}

\Rightarrow 90\times 10^{12}\le \dfrac{2^n-1}{1}

\Rightarrow 90\times 10^{12}\le 2^n-1

\Rightarrow 2^n\ge 90\times 10^{12}+1

\Rightarrow \log 2^n\ge \log (90\times 10^{12}+1)

\Rightarrow n\times \log 2\ge \log (90\times 10^{12}+1)

\Rightarrow n \ge \dfrac{\log (90\times 10^{12}+1)}{\log 2}

\Rightarrow n \ge 46.4

\Rightarrow n\approx 47


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