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Ilia_Sergeevich [38]
2 years ago
6

PLEASE HELP!!!

Mathematics
1 answer:
Svetllana [295]2 years ago
6 0

Answer:

Building linear equations for f and g, it is found that the y-intercept of (f - g)(x) is of y = 8.------------A linear function has the following format:[tex]y ...

Step-by-step explanation:Use the two points to compute the slope, m, then use one of the points in the form y=m(x)+b to find the value of b.

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Find area plsssssssssssssssss ty
natka813 [3]

Answer:

1,200

Step-by-step explanation:

20 X 12 X 5 = 1,200

7 0
3 years ago
Read 2 more answers
Find the break-even point for the given cost and revenue equations. Round to the nearest whole unit. C = 20n + 134,000 R = 160n
Harman [31]
At break-even C = R so we have:-

20n + 134,000 = 160n

140n = 134,000

answer = 134,000 / 140  =  957
8 0
3 years ago
A class quiz has 5 true or false questions the correct answers are t,t,f,f,
raketka [301]
I’m not sure what you are trying to say in this problem?
8 0
2 years ago
Which best describes the graph of g(x) = 1/9 sq x<br> PICTURE DOWN BELOW
Alexus [3.1K]

Answer:

A) The graph of g(x) is shrunk vertically by a factor of 1/9

Step-by-step explanation:

The given graph g(x) = 1/9√x

f(x) = √x

The altitude of g(x) = 1/9

The altitude of f(X) = 1

When comparing the graph of g(x) with f(x), the graph of g(x) shrunk by 1/9 because of the altitude.

Here with I have attached the graph.

Therefore, answer: A) The graph of g(x) is shrunk vertically by a factor of 1/9

Thank you.

8 0
3 years ago
AYUDA CON ESTO!!! ALGUIEN PORFAVOR
Gre4nikov [31]

Answer:

Problem 1)  frequency:  160 heartbeats per minute, period= 0.00625 minutes (or 0.375 seconds)

Problem 2) Runner B has the smallest period

Problem 3) The sound propagates faster via a solid than via air, then the sound of the train will arrive faster via the rails.

Step-by-step explanation:

The frequency of the football player is 160 heartbeats per minute.

The period is (using the equation you showed above):

Period = \frac{1}{frequency} = \frac{1}{160} \,minutes= 0.00625\,\,minutes = 0.375\,\,seconds

second problem:

Runner A does 200 loops in 60 minutes so his frequency is:

\frac{200}{60} = \frac{10}{3} \approx  3.33   loops per minute

then the period is: 0.3 minutes (does one loop in 0.3 minutes)

the other runner does 200 loops in 65 minutes, so his frequency is:

\frac{200}{65} = \frac{40}{13} \approx  3.08   loops per minute

then the period is:

\frac{13}{40} =0.325\,\,\,minutes

Therefore runner B has the smaller period

8 0
3 years ago
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