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grigory [225]
3 years ago
6

Select the correct answer.

Mathematics
2 answers:
vekshin13 years ago
4 0
The answer is d 27 games
makvit [3.9K]3 years ago
3 0
Yes, to confirm the answer is option D, 27. I got the same calculations as well. Hope this helps!
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A company manufactures two products. Market research and available resources require the following
Alex Ar [27]
Ok so basically you always don’t wanna forget to divide after multiplying and product it into A TO B not B TO A
5 0
3 years ago
Watch help
babymother [125]

Answer:

m∠RPQ = 8°

Step-by-step explanation:

m∠QRS = 4x - 15

m∠RPQ = x + 1

m∠PQR = x - 2

m∠QRS is exterior angle and m∠RPQ and m∠PQr are opposite interior angles to m∠QRS

m∠QRS = m∠RPQ + m∠PQR      {Exterior angle property of triangle}

4x - 15 = x +1 + x - 2

4x - 15 = x + x + 1-2          {Combine like terms}

4x - 15 = 2x - 1        {Subtract 2x from both sides}

4x - 2x - 15 = - 1

2x - 15 = - 1                  {Add 15 to both sides}

2x = -1 + 15

2x = 14          {Divide both sides by 2}

x = 14/2

x = 7

m∠RPQ = x + 1 = 7 + 1 = 8°

4 0
3 years ago
Read 2 more answers
What fraction of 1/2 square centimeter is 1/2 centimeter square
Brut [27]

Answer:

Step-by-step explanation

5 0
2 years ago
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Can someone pls explain!
zzz [600]

Answer: 70% is shaded. 7 are shaded and there is a total of 10. then you divide the answer would be 0.7 which is the same as 70%

Step-by-step explanation:

7 0
3 years ago
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A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min
sesenic [268]
A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})
\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}
A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4

We're given that A(0)=30. Multiply both sides by the integrating factor e^{t/100}, then

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4
\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4
e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C
A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}

Given that A(0)=30, we have

30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02

so the amount of salt in the tank at time t is

A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}
3 0
3 years ago
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