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densk [106]
3 years ago
7

A store has shirts on display. Seven of the shirts are red. There are 5 more blue shirts than green shirts. There are 2 fewer ye

llow shirts than green shirts. If there are 28 shirts in​ all, how many shirts are there of each​ color?
Mathematics
1 answer:
Nady [450]3 years ago
4 0

Answer: green shirts = 6

Blue shirts = 11

Yellow shirts = 4

Red shirts = 7

Step-by-step explanation:

Red shirts = 7

Let green shirts be x.

There are 5 more blue shirts than green shirts. This means:

Blue shirts = x + 5

There are 2 fewer yellow shirts than green shirts. This means:

Yellow shirts = x - 2

There are 28 shirts in total

Red + blue + yellow + green = 28

7 + (x + 5) + (x - 2) + x = 28

7 + x + 5 + x - 2 + x = 28

10 + 3x = 28

3x = 28 - 10

3x = 18

x = 18/3

x = 6

There are 6 green shirts

Blue shirts = x + 5 = 6 + 5 = 11

Yellow shirts = x - 2 = 6 - 2 = 4

Red shirts = 7

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If m< LMP is 11 degrees more than m< NMP and m< NML =137, find each measure
Agata [3.3K]

First, note that for angles LMP and NMP you have

m\angle LMP+m\angle NMP=m\angle NML.

If m\angle LMP is 11^{\circ} more than m\angle MNP, then

m\angle LMP=m\angle NMP+11^{\circ}.

Now, since m\angle MNL=137^{\circ}, you have

137^{\circ}=m\angle NMP+11^{\circ}+m\angle NMP,\\ \\2m\angle NMP=137^{\circ}-11^{\circ}=126^{\circ},\\ \\m\angle NMP=63^{\circ}.

Therefore,

m\angle LMP=63^{\circ}+11^{\circ}=74^{\circ}.

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In 2011, Japan experienced an intense earthquake with a magnitude of 9.1 on the Richter scale. In 2003, Japan experienced anothe
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Answer:

The intensity of the 2011 earthquake was about 6 times the intensity of the 2003 earthquake.

Step-by-step explanation:

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Substitute in the values and divide by subtracting the exponents to find

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Your answer:

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Natasha_Volkova [10]

23 Answer:  \bold{\dfrac{29,524}{9}}

<u>Step-by-step explanation:</u>

\dfrac{1}{9}+\dfrac{1}{3}+1+...+2187\\\\\\a_1=\dfrac{1}{9}=3^{-2}\qquad r=3\qquad a_n=2187\\\\\underline{\text{Find n:}}\\a_n=a_1\cdot r^{n-1}\\2187=3^{-2}(3)^{n-1}\\2187=3^{n-3}\\3^7=3^{n-3}\\7=n-3\\10=n

\underline{\text{Find the sum:}}\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\\\S_{10}=\dfrac{\frac{1}{9}(1-3^{10})}{1-3}\\\\\\.\quad =\dfrac{1-59,049}{(9)(-2)}\\\\\\.\quad =\dfrac{-59,048}{9(-2)}\\\\\\.\quad =\large\boxed{\dfrac{29,524}{9}}

24 Answer:  \bold{\dfrac{364}{9}}

<u>Step-by-step explanation:</u>

a_1=27\qquad r=\dfrac{1}{3}\qquad n=6\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\\\S_6=\dfrac{27(1-\frac{1}{3}^6)}{1-\frac{1}{3}}\\\\\\.\quad =\dfrac{27(\frac{728}{729})}{\frac{2}{3}}\\\\\\.\quad =\dfrac{27(728)}{729}\cdot \dfrac{3}{2}\\\\\\.\quad =\large\boxed{\dfrac{364}{9}}

25 Answer:  n=7

<u>Step-by-step explanation:</u>

\{-6,\ -12,\ -24,\ ...\ \}\\\\a_1=-6\qquad r=2\qquad S_n=-762\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\\\-762=\dfrac{-6(1-2^n)}{1-2}\\\\\\-762=\dfrac{-6(1-2^n)}{-1}\\\\\\\dfrac{-762}{6}=1-2^n\\\\-127=1-2^n\\\\-128=-2^n\\\\128=2^n\\\\2^7=2^n\\\\\large\boxed{7=n}

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