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postnew [5]
4 years ago
11

Drake wants to save $750 so that he can take a class on computer analysis for cars. The class is being held on various dates ove

r the next several months. Drake is planning to take the class 6 weeks from now, so he plans to save $125 each week. Unfortunately, Drake had to take out a little money from his savings in the 3rd week. After 4 weeks, Drake has $470. He knows that he must adjust his plan in order to meet his goal. Drake came up with the following options: Option A: Stay with saving the original amount each week but take the class a week later than originally planned. Option B: Increase the amount of money he saves each week by $15 from his original plan. Which of the following is a true statement? a. Only option A will allow him to meet his goal. b. Only option B will allow him to meet his goal. c. Both options A and B will allow him to meet his goal. d. Neither option A nor option B will allow him to meet his goal.
Mathematics
1 answer:
luda_lava [24]4 years ago
3 0
Option A: He currently has $470 with 3 weeks to go (taking the class 1 week later as planned). Saving $125 each week for 3 weeks would give him a total of $845, giving him the needed money to take the classes.

Option B: He currently has $470 with 2 weeks to go. Instead of just saving up an amount of $125 per week for the remaining 2 weeks, he decides to add $15, saving up to $140 per week instead. Saving this amount for 2 weeks gives him $280, bringing his total to $750, which is enough for him to take the classes.
So, therefore, the answer is C.
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write the ratio as a percent 6 per 100 , 49 out of 100 , 16 per 100 , 21 out of 100 , 80 per 100 , 93 out of 100​
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6%, 49%, 16%, 21%, 80%, 93%

Step-by-step explanation:

add the percent sign to each each number (6, 49, 16, 21, 80, and 93).

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3 years ago
Read 2 more answers
1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a
katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

8 0
3 years ago
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