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SVETLANKA909090 [29]
3 years ago
9

On its first visit to the vet, Emma’s kitten weighed 30 ounces. On its second visit, the kitten had gained 8 ounces. On its thir

d visit, the kitten had gained another 6 ounces. What is the percent increase, rounded to the nearest percent, of the kitten’s weight since its first visit?
Mathematics
2 answers:
IrinaK [193]3 years ago
7 0

Answer:

46.667% increase of 30.

Step-by-step explanation:

44 is a 46.667% increase of 30.

PSYCHO15rus [73]3 years ago
4 0

Step-by-step explanation:

Thank you rhhtthrhrgrgtnymuge hjk it is the right

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Create a graph that has the following intercepts. Make sure to label each of the intercepts on
SCORPION-xisa [38]

Answer:

  see below for the graph

Step-by-step explanation:

The desired graph has two y-intercepts and one x-intercept. It is not the graph of a function.

Here's one way to get there.

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Start with the parent function y = |x| and scale it down so that it has a y-intercept of -1 and x-intercepts at ±1.

Now, it is ...

  f(x) = |x| -1

We want to scale this vertically by a factor of -5. this puts the y-intercept at +5 and leaves the x-intercepts at ±1.

Horizontally, we want to scale the function by an expansion factor of 3. The transformed function g(x) will be ...

  g(x) = -5f(x/3) = -5(|x/3| -1) = -5/3|x| +5

This function has two x-intercepts at ±3 and one y-intercept at y=5. By swapping the x- and y-variables, we can get an equation for the graph you want:

  x = -(5/3)|y| +5

______

<em>Comment on this answer</em>

Since there are no requirements on the graph other than it have the listed intercepts, you can draw it free-hand through the intercept points. It need not be describable by an equation.

7 0
4 years ago
10 POINTS!!! FULL ANSWER IN STEP BY STEP FORMAT!!
stich3 [128]
A) The signs of the first derivative (g') tell you the graph increases as you go left from x=4 and as you go right from x=-2. Since g(4) < g(-2), one absolute extreme is (4, g(4)) = (4, 1).

The sign of the first derivative changes at x=0, at which point the slope is undefined (the curve is vertical). The curve approaches +∞ at x=0 both from the left and from the right, so the other absolute extreme is (0, +∞).

b) The second derivative (g'') changes sign at x=2, so there is a point of inflection there.

c) There is a vertical asymptote at x=0 and a flat spot at x=2. The curve goes through the points (-2, 5) and (4, 1), is increasing to the left of x=0 and non-increasing to the right of x=0. The curve is concave upward on [-2, 0) and (0, 2) and concave downward on (2, 4]. A possible graph is shown, along with the first and second derivatives.

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3 years ago
I'm trying to do this problem, but I have no idea how to even start it:
zhannawk [14.2K]

Answer given in the picture above.

3 0
4 years ago
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jok3333 [9.3K]
Simple..you get the RAtional answer of...

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d1i1m1o1n [39]

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