The correct answers are B and C.

Since the mixture requires 5 mL of black wax and 6 mL of white wax, we want multiples of these numbers; specifically, we need multiples that are the same number of times away.

A does not work; 10 is 5*2 and 24 is 6*4. These won't be in the same ratio.

B works; 15 is 5*3 and 18 is 6*3. These have the same ratio.

C works; 35 is 5*7 and 42 is 6*7. These have the same ratio.

D does not work; 26 is not a multiple of 6.

E does not work; 45 is 5*9 and 48 is 6*8. These will not have the same ratio.

8 0

2 years ago

Read 2 more answers

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 (if $A^{\prime\prime}(t) = 0$ ) 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 missing angle measure of the triangle

Tomtit [17]

Answer: 45°

Step-by-step explanation: add the angles together to get a total of 180°

6 0

2 years ago