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Serjik [45]
3 years ago
8

Let f(x)=x^2-3x+5. Find the average rate of change from x=-3 to x=0

Mathematics
1 answer:
Ann [662]3 years ago
5 0

Answer:

Step-by-step explanation:

Differentiate

2x-3

Subsituite values of x given

2(0)-3 - 2(-3) - 3

0

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Robert bought $1.60 worth of stamps with 20 coins all in nickels and dimes. how many nickels and dimes did he use?
mrs_skeptik [129]
Nickels are 5 cents, and dimes are 10 cents.
(By luck..) eventually you can have 12 dimes, meaning you have 120 cents, or $1.20 ... 20-12=8.. meaning you used 12 dimes and are missing 8 more coins. which means you can have 8 nickels which is 40 cents. 120+40=160 cents or $1.60
Hope this helped.

3 0
3 years ago
Câu 5 (3.0 điểm) Cho đường tròn (O) bán kính R ngoại tiếp tam giác ABC có ba góc nhọn.
Luda [366]

Answer:

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2 years ago
The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s
oksian1 [2.3K]

Answer:

The maximum acceleration over that interval is A(6) = 28.

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable t.

Notice that the interval of interest 0 \le t \le 6 is closed on both ends. In other words, this interval includes both endpoints: t = 0 and t= 6. Over this interval, the value of A(t) might be maximized when t is at the following:

  • One of the two endpoints of this interval, where t = 0 or t = 6.
  • A local maximum of A(t), where A^\prime(t) = 0 (first derivative of A(t)\! is zero) and A^{\prime\prime}(t) (second derivative of \! A(t) is smaller than zero.)

Start by calculating the value of A(t) at the two endpoints:

  • A(0) = 10.
  • A(6) = 28.

Apply the power rule to find the first and second derivatives of A(t):

\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}.

\displaystyle A^{\prime\prime}(t) = 6\, t - 15.

Notice that both t = 1 and t = 4 are first derivatives of A^{\prime}(t) over the interval 0 \le t \le 6.

However, among these two zeros, only t = 1\! ensures that the second derivative A^{\prime\prime}(t) is smaller than zero (that is: A^{\prime\prime}(1) < 0.) If the second derivative A^{\prime\prime}(t)\! is non-negative, that zero of A^{\prime}(t) would either be an inflection point (ifA^{\prime\prime}(t) = 0) or a local minimum (if A^{\prime\prime}(t) > 0.)

Therefore \! t = 1 would be the only local maximum over the interval 0 \le t \le 6\!.

Calculate the value of A(t) at this local maximum:

  • A(1) = 15.5.

Compare these three possible maximum values of A(t) over the interval 0 \le t \le 6. Apparently, t = 6 would maximize the value of A(t)\!. That is: A(6) = 28 gives the maximum value of \! A(t) over the interval 0 \le t \le 6\!.

However, note that the maximum over this interval exists because t = 6\! is indeed part of the 0 \le t \le 6 interval. For example, the same A(t) would have no maximum over the interval 0 \le t < 6 (which does not include t = 6.)

4 0
3 years ago
Rylie's gross paycheck amount is $1305.60. She has 4% deducted from her paychecks for her 401(k). Her employer matches her deduc
vaieri [72.5K]
The employer contribution does not get deducted from her paycheck.  To find the amount deducted from her check we take 4% of 1305.60 = 0.04(1305.60) = $52.22.
6 0
3 years ago
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Does this graph represent a function ? Why or why not ?
balu736 [363]

Answer:

D) It passes the horizontal line test

Step-by-step explanation:

In order for a line to represent a function it must pass the vertical line test.

To apply the horizontal line test, take a pencil and hold it vertically. Then, move it across the graph from left to right. If it intersects the line at 1 point everywhere along the line, then it is a function (functions can only have 1 Y value for every X value).

3 0
3 years ago
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