Solve for s. s 4 − 2 = 2 s =

Mark and peter went to an arcade where the machines took tokens. Marilk played 9 games of ping pong and 5 games of pinball, usin

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Answer:

9x + 5y = 29 ........... (1) and

3x + y = 7 ........... (2)

Each game of ping pong requires 1 number of token and each game of pinball requires 4 number of tokens.

Step-by-step explanation:

Let, each game of ping pong requires x number of tokens and each game of pinball requires y number of tokens.

So, from the given conditions we can write

9x + 5y = 29 ........... (1) and

3x + y = 7 ........... (2)

Now, solving equations (1) and (2) we get,

9x + 5(7 - 3x) = 29

⇒ 35 - 6x = 29

⇒ 6x = 6

⇒ x = 1 token.

Now, putting x = 1 in equation (2) we get,

3 + y = 7

⇒ y = 4 tokens.

So, each game of ping pong requires 1 number of token and each game of pinball requires 4 number of tokens. (Answer)

6 0

3 years ago

Which of the following are possible solutions to following system of linear inequalities:

lapo4ka [179]

The answer Is F or the C

5 0

3 years ago

The population, P(t), of China, in billions, can be approximated by1 P(t)=1.394(1.006)t, where t is the number of years since th

vitfil [10]

Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

Step-by-step explanation:

To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

We know that the population, P(t), of China, in billions, can be approximated by $P(t)=1.394(1.006)^t$

To find the derivative you need to:

$\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t$

To find the population growing at the start of 2014 we say t = 0

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year$

To find the population growing at the start of 2015 we say t = 1

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year$