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Dovator [93]
2 years ago
14

What is the cheapest way to go to a ball game if the admissions options are as shown? A) $10 for a single B) $15 for a couple C)

$20 for a triple D) $35 for a group of four E) $50 for a group of five
Mathematics
1 answer:
vivado [14]2 years ago
5 0

Answer: C

Step-by-step explanation: A = 10 for one person. B = 7.5 for one person. C = 6.67 for one person. D = 8.75 for one person. E = 10 for one person.

Out of all of these, C is the cheapest.

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Please help me if you can. Please also write how you got the answer thank you.
vaieri [72.5K]

Answer:

The three numbers are 341, 342, and 343

Step-by-step explanation:

We start by assigning X to the first integer. Since they are consecutive, it means that the 2nd number will be X + 1 and the 3rd number will be X + 2 and they should all add up to 1026. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 1026

To solve for X, you first add the integers together and the X variables together. Then you subtract three from each side, followed by dividing by 3 on each side. Here is the work to show our math:

X + X + 1 + X + 2 = 1026

3X + 3 = 1026

3X + 3 - 3 = 1026 - 3

3X = 1023

3X/3 = 1023/3

X = 341

Which means that the first number is 341, the second number is 341 + 1 and the third number is 341 + 2. Therefore, three consecutive integers that add up to 1026 are 341, 342, and 343.

341 + 342 + 343 = 1026

We know our answer is correct because 341 + 342 + 343 equals 1026 as displayed above.

6 0
2 years ago
Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{
SVETLANKA909090 [29]

y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

-----------------------

  • Note 1: A more general solution is y = Ce^x for some constant C.
  • Note 2: It might be tempting to say the general solution is y = e^x+C, but that is not the case because y = e^x+C \to y' = e^x+0 = e^x and we can see that y' = y would only be true for C = 0, so that is why y = e^x+C does not work.
6 0
2 years ago
Evaluate the expression
zheka24 [161]

Answer:

Step-by-step explanation:

x³ = 4³ = 4 * 4 * 4 = 64

4 0
2 years ago
Read 2 more answers
PLEASE ANSWER
gladu [14]

Answer:

420 + 15 = 435

360 + 45 = 405

435 - 405 = 30.

30 minutes less sunlight

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
675 grams of sweets were distributed among 9 children. How much sweet was given to each child?
NISA [10]

Answer:

75 Sweets

Step-by-step explanation:

To get the answer you have to divide the number of sweets by the number of children. If there are 675 grams and 9 children, then you divide 675/9 to get your answer of 75 sweets.

8 0
2 years ago
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