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nataly862011 [7]
4 years ago
11

Giselle enjoys playing chess and Scrabble. She wants to play a total of at least 10 games (condition A), but she has only 6 hour

s to do it (condition B)
The graph represents the constraints on the number of chess games C and Scrabble games S Giselle plays.
What is the greatest number of chess games Giselle can play while meeting both constraints?

Mathematics
1 answer:
cestrela7 [59]4 years ago
8 0

Answer:

She only can play at most 4 chess games

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Divide the first equation by 2 and add the result to the second equation. This will eliminate x.

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... y = -6 . . . . . . . divide by 3

Substitute this into either equation to find x. Let's use the second equation, where the coefficient of x is positive.

... 2x +4(-6) = -12

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The solution is (x, y) = (6, -6).

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James is a welder who earns $20 per hour. He works 8 hours per day, Monday through Friday. How much does James earn in one week?
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The slope is -2. The y-intercept is (0,1).
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The surface area of a cylinder is increasing at a rate of 9 pi square meters per hour. The height of the cylinder is fixed at 3
Alekssandra [29.7K]

Answer:

9\pi \text{ cubic meters per hour}

Step-by-step explanation:

Since, the surface area of a cylinder,

A= 2\pi r^2 + 2\pi rh  ................(1)

Where,

r = radius,

h = height,

If A= 36\pi\text{ square meters}, h = 3\text{ meters}

36\pi = 2\pi r^2 + 2\pi r(3)

18 = r^2 + 3r

\implies r^2 + 3r - 18=0

r^2 + 6r - 3r - 18 = 0     ( by middle term splitting )

r(r+6)-3(r+6)=0

(r-3)(r+6)=0

By zero product property,

r = 3 or r = - 6 ( not possible )

Thus, radius, r = 3 meters,

Now, differentiating equation (1) with respect to t ( time ),

\frac{dA}{dt}= 4\pi r\frac{dr}{dt} +2\pi(r\frac{dh}{dt} + h\frac{dr}{dt})

∵ h = constant, ⇒ dh/dt = 0,

\frac{dA}{dt} = 4\pi r \frac{dr}{dt} +2\pi h \frac{dr}{dt}

We have, \frac{dA}{dt}=9\pi\text{ square meters per hour}, r = h = 3\text{ meters}

9\pi = 4\pi (3) \frac{dr}{dt}+2\pi (3)\frac{dr}{dt}

9\pi = (12\pi + 6\pi )\frac{dr}{dt}

9\pi = 18\pi \frac{dr}{dt}

\implies \frac{dr}{dt} =\frac{1}{2}\text{ meter per hour}

Now,

Volume of a cylinder,

V=\pi r^2 h

Differentiating w. r. t. t,

\frac{dV}{dt}=\pi ( r^2 \frac{dh}{dt}+h(2r)\frac{dr}{dt})=\pi ((3)(6) (\frac{1}{2})) = 9\pi \text{ cubic meters per hour}

6 0
3 years ago
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