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Y_Kistochka [10]
3 years ago
9

Edward is making a rectangular picture frame. He wants the perimeter of a frame to be no more than 96 inches. He also wants the

length of the frame to be greater than or equal to the square of 4 inches less than its width.
Instructions: Create a system of inequalities tomorrow love you both situation and use it to determine how many of the solutions are viable.
Mathematics
1 answer:
Bess [88]3 years ago
8 0

No more than in terms of math means less than or equal to.

P = 2l + 2w

{96 ≤ 2l + 2w

{L ≥ w - 4

Here we see a system of linear inequality in two variables.

You finish.

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2 years ago
What value of n makes the equation true? (2x^9y^n)(4x^2y^10)=8x^11y^20
Komok [63]
Hi there! The answer is n = 10.

(2 {x}^{9}  {y}^{n} )(4 {x}^{2}  {y}^{10} ) = 8 {x}^{11}  {y}^{20}

As you see at the powers of x, we need to add the exponents of the power we when multiply them.
{x}^{9}  \times  {x}^{2}  =  {x}^{9 + 2}  = x {}^{11}

The powers of y work the same way.
{y}^{n}  \times  {y}^{10}  =  {y}^{n + 10}  =  {y}^{20}

Hence, n = 10, since
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8 0
3 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
Find the value of A that makes the following equation true for all values of x. (1/7)^x = A^-9x.
elena-14-01-66 [18.8K]

Answer:

D

Step-by-step explanation:

(1/7)^(x)=A^(-9x)

(1/7)^(x)=(A^(-9))^x

A^(-9)=1/7

A=(1/7)^9

3 0
2 years ago
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