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maksim [4K]
2 years ago
11

Your next math test is worth 111 points and contains 32 problems. Each problem is worth either 4 points or 3 points. How many 3

point problems are on the test?
Mathematics
2 answers:
faust18 [17]2 years ago
5 0

Answer:

usatestprep?

Step-by-step explanation:

Dafna11 [192]2 years ago
3 0
We call x the 4 points-worth problems and y the 3 points-worth problems

You know that x+y = 32

4x+3y = 111

You know that the difference from the x and y is 32 so write:

x = 32-y

Substitute at x the value of 32-y

4(32-y)+3y = 111

128-4y +3y = 111

-4y+3y = 111-128

-y = -17

y = 17
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\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
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\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
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\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
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\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
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