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Gennadij [26K]
4 years ago
14

Which of the following represents a function with the greatest rate of change?

Mathematics
2 answers:
pshichka [43]4 years ago
6 0
It is C because when you calculate you can see I it has more change
defon4 years ago
5 0
So the rate of change can be found in different places.

For the first one:

This equation is in slope-intercept form, or y = mx + b.

m would be the slope/rate of change, and b the y intercept. (x and y would just be any sets of ordered pairs that fit into the equation.)

So if you look at which number is in the m spot, or the rate of change spot, you see that 5 is the rate of change.

For the second one: The formula for slope is:

m = \frac{ y_{2} -  y_{1} }{x_{2} -  x_{1} }

So just put in any ordered pairs in. I'm just going to put in the ordered pairs, (6, 30) and (4, 22).

Or,

m = (30 - 22) / (6 - 4)

Or,

m = 8/2

Or 4.

For the third one:

You can use the same formula as I stated above.

So you have to look for two ordered pairs here, two points.

So you can estimate that two of the points are: (0, 3) and (-2, -9).

So then just solve for that.

(-9 - 3) / (-2 - 0) = -12 / -2 = 6

For the last one:

Again, use the same equation.

m = (7 - 1) / (-1 - (-3))

 m = 6/2 

m = 3

So the one with the greatest rate of change would be C.
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The equation of the circle is may be expressed as, 
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where h and k are the abscissa and ordinates of its center, respectively, and r is radius. Substituting the given values to the equation above, 
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Original Price $60 Percent of Discount Sale Price $45
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Answer:

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Step-by-step explanation:

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3 years ago
Can someone help please:<br> let f(x)=x^2+x <br> solve f(x-1)+f(x+1)=6
eimsori [14]

Step-by-step explanation:

f(x)=x^2+x

so

f(x-1) = (x - 1) ^ 2 + (x - 1)

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= x^2 - 3x

and

f(x+1) = (x + 1) ^ 2 + (x + 1)

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= x^2 + 3x + 2

now

f(x-1) + f(x+1) = 6

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2x^2 = 4

X = √2

brainliest pls took long enough to solve

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Step-by-step explanation:

ondclqw/mcnkw;oj'cpml icojn jhlcihojwlnjkjhbiuhpojwklnejhjclbiuqhojwk/ne clbiowjelcl jhqbwio;elc bqlhjwieo;jclnjl qjwleipc'lq wljbepcijoqlnwjc

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