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qwelly [4]
3 years ago
14

Durant and Westbrook are playing a three-point challenge game to improve long distance shooting. Each time each of them makes a

basket, 3 points are added to their respective scores. Durant has 18 points and Westbrook has 12 points.
Which of the following options has the pair of explicit formulas that represents the number of points for each player as an arithmetic sequence?

Durant: 18+(n-1)(3) Westbrook: 12+(n-1)(3)
Durant: 18-(n-1)(3) Westbrook: 12-(n-1)(3)
Durant: 18+(n-3) Westbrook: 12+(n-3)
Durant: 18+3n Westbrook: 12+3n
Mathematics
1 answer:
Tasya [4]3 years ago
6 0

Answer:The answer is D

Step-by-step explanation:

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Round the nine to the nearest hundred, (947.362)
DochEvi [55]

Answer:

900?

Step-by-step explanation:

.... self explanatory

6 0
3 years ago
I just need to confirm if I am correct
Vikentia [17]
Yup. the x axis is the base and the y axis is the height, so it’s right.
5 0
3 years ago
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Coach Stevens needs to purchase sprinklers to water the baseball field. The standard distance between bases is 90 feet and the i
egoroff_w [7]
If you need the geometry and trigonometry I'll post those.  Instead I found an answer at this link: http://jwilson.coe.uga.edu/EMAT6450/Class%20Projects/Scarpelli/Scarpelli_MathematicsBaseballActivity.....
The distance from home plate to the pitcher's mound is 60.5 feet and from home plate to second base is <span>127.28 feet.
Pitcher's Mound to 2nd base = </span><span> <span> <span> 66.78</span> </span>f</span>eet

Given those distances, we KNOW a 50 foot sprinkler will NOT reach home plate and second base from the pitcher's mound.

I didn't figure out the pitcher's mound to 1st or to 3rd, since the question is already answered.


6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5C%5C%20%5Cleq%20%5Cbeta%20%5Calpha%20%5Calpha" id="TexFormula1" title="\\ \leq \beta \alpha
nadya68 [22]
Hi lol I don’t understand you’re question I need more context
8 0
3 years ago
How to solve an expression with variables in the exponents?
gtnhenbr [62]

Answer:

  use logarithms

Step-by-step explanation:

Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.

__

You will note that this approach works well enough for ...

  a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents

  (x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs

but doesn't do anything to help you solve ...

  x +3 = b^(x -6)

There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.

__

Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.

In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.

8 0
3 years ago
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