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Svet_ta [14]
3 years ago
12

The Tres Difficult race helps raise money for charity. According to the website, of the proceeds from ticket sales go directly t

o charity.
2/5
Last year they made $8000 from ticket sales. How much was given to charity?
Mathematics
2 answers:
DedPeter [7]3 years ago
8 0

Answer:

3200

Step-by-step explanation:

We need to find 2/5 of the tickets sales

2/5 * 8000

3200

Gre4nikov [31]3 years ago
7 0

Answer:

3200

Step-by-step explanation:

you need to find what 2/5 is and the you take that away from 8000 and then you have your answer of 3200

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3 years ago
Find the general indefinite integral. (Use C for the constant of integration. Remember to use absolute values where appropriate.
Maru [420]

Answer:

9\text{ln}|x|+2\sqrt{x}+x+C

Step-by-step explanation:

We have been an integral \int \frac{9+\sqrt{x}+x}{x}dx. We are asked to find the general solution for the given indefinite integral.

We can rewrite our given integral as:

\int \frac{9}{x}+\frac{\sqrt{x}}{x}+\frac{x}{x}dx

\int \frac{9}{x}+\frac{1}{\sqrt{x}}+1dx

Now, we will apply the sum rule of integrals as:

\int \frac{9}{x}dx+\int \frac{1}{\sqrt{x}}dx+\int 1dx

9\int \frac{1}{x}dx+\int x^{-\frac{1}{2}}dx+\int 1dx

Using common integral \int \frac{1}{x}dx=\text{ln}|x|, we will get:

9\text{ln}|x|+\int x^{-\frac{1}{2}}dx+\int 1dx

Now, we will use power rule of integrals as:

9\text{ln}|x|+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\int 1dx

9\text{ln}|x|+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+\int 1dx

9\text{ln}|x|+2x^{\frac{1}{2}}+\int 1dx

9\text{ln}|x|+2\sqrt{x}+\int 1dx

We know that integral of a constant is equal to constant times x, so integral of 1 would be x.

9\text{ln}|x|+2\sqrt{x}+x+C

Therefore, our required integral would be 9\text{ln}|x|+2\sqrt{x}+x+C.

4 0
3 years ago
Enter an algebraic expression that represents the difference between e squared and twenty-one
Readme [11.4K]

Answer: e^2 - 21

Step-by-step explanation:

e squared will  e^2  

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The diffrence is the same as subtraction .  

e^2 - 12

3 0
3 years ago
You need to invest $1000 in a bank account and are give two options. The first option is to earn $50 every month you leave the m
garri49 [273]

Answer:

<h3><u>Option 1</u></h3>

Earn $50 every month.

  • Let x = number of months the money is left in the account
  • Let y = the amount in the account
  • Initial amount = $1,000

\implies y = 50x + 1000

This is a <u>linear function</u>.

<h3><u>Option 2</u></h3>

Earn 3% interest each month.

(Assuming the interest earned each month is <u>compounding interest</u>.)

  • Let x = number of months the money is left in the account
  • Let y = the amount in the account
  • Initial amount = $1,000

\implies y = 1000(1.03)^x

This is an <u>exponential function</u>.

<h3><u>Table of values</u></h3>

<u />

\large \begin{array}{| c | l | l |}\cline{1-3} & \multicolumn{2}{|c|}{\sf Account\:Balance} \\ \cline{1-3} & \sf Option\:1 & \sf Option\:2 \\\sf Month & \sf \$50\:per\:mth & \sf 3\%\:per\:mth \\\cline{1-3} 0 & \$1000 & \$1000 \\\cline{1-3} 1 & \$1050 & \$1030 \\\cline{1-3} 2 & \$1100 & \$1060.90 \\\cline{1-3} 3 & \$1150 & \$1092.73 \\\cline{1-3} 4 & \$1200 & \$1125.51 \\\cline{1-3} 5 & \$1250 & \$1159.27 \\\cline{1-3} 6 & \$1300 & \$1194.05 \\\cline{1-3} 7 & \$1350 & \$1229.87 \\\cline{1-3}\end{array}

From the table of values, it appears that <u>Account Option 1</u> is the best choice, as the accumulative growth of this account is higher than the other account option.

However, there will be a point in time when Account Option 2 starts accruing more than Account Option 2 each month.  To find this, graph the two functions and find the <u>point of intersection</u>.

From the attached graph, Account Option 1 accrues more until month 32.  From month 33, Account Option 2 accrues more in the account.

<h3><u>Conclusion</u></h3>

If the money is going to be invested for less than 33 months then Account Option 1 is the better choice.  However, if the money is going to be invested for 33 months or more, then Account Option 2 is the better choice.

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Therefore, set y = 7
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Solve for x
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3 years ago
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