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horrorfan [7]
3 years ago
7

Hey please help, thank you!!

Mathematics
1 answer:
marin [14]3 years ago
5 0

Answer:

D

Step-by-step explanation:

So we have the equation:

A=\frac{1}{2}(12)(3+7)

And we want to find out which of the answer choices is <em>not</em> equivalent.

Thus, let's go through each of the choices.

A)

We have:

\frac{12(3+7)}{2}

This is correct.

Our original equation can be written as:

A=\frac{1}{2}(\frac{12(3+7)}{1})

And if we multiply straight across:

A=\frac{12(3+7)}{2}

We'll get choice A. So choice A is correct.

B)

We have:

6(3+7)

This is also correct.

From our original equation:

A=\frac{1}{2}(12)\cdot(3+7)

If we multiply the first term together:

A=6(3+7)

We'll get choice B. So, choice B is correct.

C)

We have:

0.5(12)(10)

This is correct as well.

If we reduce 1/2 to decimal form and add the operations within the parentheses in the original equation, we will get:

A=\frac{1}{2}(12)(3+7)\\A=0.5(12)(10)

So, C is also correct.

D)

We have:

\frac{12}{2}\times\frac{10}{2}

Note that this is not correct.

This attempted to use the distributive property. However, the distribute property does <em>not work</em> for multiplication. For instance:

3(2+1)\text{ indeed equals } 3(2)+3(1)

However:

3(2\times1)\text{ does not equal} 3(2)\times3(1)

So, choice D is not equivalent.

So choice D is our answer :)

If we evaluate it, we will get 30, in contrast to the 60 of the others.

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Answer:

70 dolls

Step-by-step explanation:

Hello!

The ratio is 5:2, which are division of  a whole. They can be represented as 5x and 2x.

Jacky has 5x dolls, and Peter has 2x dolls. They were equal after subtracting 15 dolls from 5x and adding them to 2x.

Equation:

  • 5x - 15 = 2x + 15

Solve:

  • 5x - 15 = 2x + 15
  • 5x - 30 = 2x
  • 5x = 2x + 30
  • 3x = 30
  • x = 10

So Jacky has 5(10) dolls, or 50 dolls, and Peter has 2(10) or 20 dolls.

The total sum is 70 dolls.

4 0
2 years ago
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Therefore
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Theo deposits $2,000 deposit in a savings account earning compound interest at an annual rate of 5% compounded annually. He make
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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.)

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