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Ierofanga [76]
3 years ago
6

I’ll give brainiest :) pls help ASAP

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
7 0

Answer:

10⁷ is the correct answer of the question

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PLEASE HELP ME! A skydiver parachutes to the ground. The height y (in feet) of the skydiver after x seconds is y = -10x + 3000.
tatuchka [14]
A) Plot a point at (0, 3000), then subtract 10 for every second that passes by.
B) The slope is negative because the skydiver is falling and the graph is plotting his descent, his x intercept is at (300, 0) which means he will touch land after 300 seconds
To find the the x intercept, make y equal 0 so 0=-10x+3000, subtract by 3000 on each side, and you end up with -3000=-10x, you then divide both sides by -10 and you end up 300=x or x=300
4 0
3 years ago
Read 2 more answers
Simplify. 4a(6) + 3a
Iteru [2.4K]

Answer: 27a

Step-by-step explanation:

1. Simplify 4a×6 to 24a.

24a+3a

2. Simplify.

27a

8 0
3 years ago
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Sarah and Annie brought some quarters to an arcade. Sarah has 28 more quarters than Annie.
Marina CMI [18]

Answer:

37

Step-by-step explanation:

m=28+n

36=28+n

n=36-28

n=8

Annie quarters:

29+n

29+8

37

5 0
3 years ago
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Given the following linear functio, determine the slope of a line parallel to f(x)
andrew11 [14]

Answer:

Option C is correct.

C. 4/3

Step-by-step explanation:

Given:

The given linear function of a line.

F(x)=4/3x+2

We need to determine the slope of a line parallel to F(x)=4/3x+2

Solution:

First we compare the given function of the line with standard form of a line. y = mx + c

Where m = slope of the line.

And c = y-intercept

So, the slope of the line m=\frac{4}{3} and y-intercept c=2

We know that the slope of the parallel lines are same.

Therefore, the slope of a the parallel line m=\frac{4}{3}.

8 0
3 years ago
Find the sum of series 4/(4n-3) (4n+1)
ANTONII [103]
\dfrac4{(4n-3)(4n+1)}=\dfrac1{4n-3}-\dfrac1{4n-1}

Assuming the sum starts at n=1, the Nth partial sum is

\displaystyle\sum_{n=1}^N\left(\frac1{4n-3}-\frac1{4n-1}\right)=\left(1-\frac15\right)+\left(\frac15-\frac19\right)+\cdots+\left(\frac1{4N-7}-\frac1{4N-3}\right)+\left(\frac1{4N-3}-\frac1{4N+1}\right)
\displaystyle\sum_{n=1}^N\left(\frac1{4n-3}-\frac1{4n-1}\right)=1-\frac1{4N-1}

As N\to\infty, you're left with simply 1.
8 0
3 years ago
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