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pogonyaev
3 years ago
11

The first group was responsible for making 160 suits while the second group was responsible for making 25% fewer suits in the sa

me amount of time. The first team made 10 more suits each day than the second team and finished its job 2 days before the deadline date. How many suits did the second team make each day if it needed two days extra after the deadline to finish the assigned job?
Mathematics
1 answer:
egoroff_w [7]3 years ago
3 0
First group rate is x+10, Second is x
First group made 160, Second made 120
Time the first group will take is 160/x+10, and second is 120/x
Equation is: 160/x+10 + 4 = 120/x
x=10, x=-30
The second group made 10 suits everyday.


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Radda [10]

Answer:

Angle sqr = 119°

Step-by-step explanation:

In the above question, we are given:

angle pqt is (3x+47)°

angle sqr is (6x-25)°

Angle pqt = Angle sat

We solve for x

3x + 47 = 6x - 25

Collect like terms

47 + 25 = 6x - 3x

72 = 3x

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We solve for the measure of angle sqr

Angle sqr = (6x - 25)

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= 119°

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3 years ago
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A construction crew is lengthening a road. The road started with a length of 51 miles, and the crew is adding 2 miles to the roa
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Total length after 33 days will be 117 miles

Step-by-step explanation:

A construction crew is lengthening a road. The road started with a length of 51 miles.

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5 0
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Evaluate the sum of the following finite geometric series.
rjkz [21]

Answer:

\large\boxed{\dfrac{156}{125}\approx1.2}

Step-by-step explanation:

<h3>Method 1:</h3>

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}

<h3>Method 2:</h3>

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}

a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1

\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}

5 0
3 years ago
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