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UkoKoshka [18]
4 years ago
5

A bowling alley has a fixed base cost and charges a variable per game rate. It costs $20.50 for five games and $28.50 for nine g

ames, what is the variable cost
Mathematics
1 answer:
tiny-mole [99]4 years ago
5 0
Answer: 2
 Work: Substract 28.5 minus 20.5 = 8 Then substract 9 - 5 = 4
Then put them as a fraction like 8/4
Put it as a unit rate or divide 8 by 4 and that equals 2. Hope it helps!
Oh and there's also a website in Google with all the answers with the work! :D
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Step-by-step explanation:

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3 years ago
Two vertical poles of heights 60 ft and 80 ft stand on level ground, with their bases 100 ft apart. A cable that is stretched fr
Andreyy89

Answer:

Step-by-step explanation:

Given

height of two Poles are 60 and 80 ft

Distance between them is 100 ft

Let x be the distance of Pole of ht 80 ft from Point of stretch

thus length of rope is given by

L=L_1+L_2

L_1=\sqrt{80^2+x^2}

L_2=\sqrt{60^2+(100-x)^2}

L=\sqrt{80^2+x^2}+\sqrt{60^2+(100-x)^2}

differentiate w.r.t x we get

\frac{\mathrm{d} L}{\mathrm{d} x}=\frac{2x}{2\sqrt{80^2+x^2}}-\frac{2\left ( 100-x\right )}{\sqrt{60^2+\left ( 100-2x\right )^2}}

Put \frac{\mathrm{d} L}{\mathrm{d} x}=0 to get minimum value

\frac{2x}{2\sqrt{80^2+x^2}}=\frac{2\left ( 100-x\right )}{\sqrt{60^2+\left ( 100-2x\right )^2}}

squaring

x^2\left ( (100-x)^2+60^2\right )=(100-x)^2(80^2+x^2)

Rearranging

28x^2-1800x+640000=0

x=\frac{400}{7}

thus L_1=98.312 ft

L_2=73.73 ft

L=172.04 ft  

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