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lyudmila [28]
3 years ago
9

Susan is writing a linear equation for the cost of her cell phone plan. In the first month, she talks for 52 minutes and is char

ged $19.41. In the second month, she talks for 380 minutes and is charged $45.65. A) How much money does Susan's cell phone company charge for each minute? B) Write an equation that Susan can use to determine the cost of her cell phone plan, y, as it relates to the number of minutes used, x
Mathematics
1 answer:
Alex787 [66]3 years ago
5 0
Assuming that the cost per minute is the same for both months and the plan fee is the same, you can use y=mx+b for this

y is the cost of the phone plan, x is the cost per minute and b is the start cost.

so    19.41=25x+b for the first month
and  45.65=380x+b for the second month

solve both for b you get:
19.41-25x=b and 45.65-380x=b.  from this we get
19.41-25x=45.65-380x

solve for x

328x=26.24 and x=0.08

this means the cost per minute is 0.08c/min (answer A)

rewrite the equation to calculate b, and where this time, the x is the number of minutes talked.

y=0.08x+b and plug in one of the two months

45.65=0.08 * 380 + b

Solve for b and b is 15.25

so the final equation is

y=0.08x+15.25  (answer B)
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Step-by-step explanation:

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All integrals in the form \int\limits {\sqrt{a^{2} -x^{2} } } \, dx are always evaluated using the substitute given where 'a' is any constant.

From the given integral, \int\limits {7\sqrt{49-x^{2} } } \, dx = \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx where a = 7 in this case.

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\int\limits {7\sqrt{49-x^{2} } } \, dx \\= \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -(7sin\theta)^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -49sin^{2}\theta  } } \, dx\\= \int\limits {7\sqrt{49(1-sin^{2}\theta)}   } } \, dx\\= \int\limits {7\sqrt{49(cos^{2}\theta)}   } } \, dx\\since\ dx = 7cos\theta d\theta\\= \int\limits {7\sqrt{49(cos^{2}\theta)}   } } \, 7cos\theta d\theta\\= \int\limits {7\{7(cos\theta)}   }}} \, 7cos\theta d\theta\\

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Answer:

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