Write a rule for each arithmetic sequence -3, -0.5, 2, 4.5,....

If the domain of the square root function f(x) is x

soldier1979 [14.2K]

<span>This question, in my opinion, is not well stated. If f(x) = √x, as the question statement seems to say, then the domain is not x<7. Rather, the domain is x≥0. If f(x) is not the square root function, but say f(x) = √(7-x) then the domain is x≤7, and for this function then the appropriate answer is d), since the x-term inside the radical has a negative coefficient.</span>

4 0

3 years ago

Because bernard has some health issues, he must pay 15% more for life insurance. about how much more annually will a $115,000 10

Sergeu [11.5K]

The amount more annually a $115,000 10-year term insurance at age 35 cost Bernard than someone of the same age without health issues is $24.

<h3>What are insurance premiums?</h3>

The insurance premium is paid as a cost to cover a possible loss that is unseen.

The annual premium rate as a percentage of the value insured a person at age 35 has to pay is 0.14%.

From the given information, we have that the amount a 35-year-old without health issues will pay per $1,000 is $1.40

The amount to be paid for $115,000 is 115 × $1.4 = $161

The amount Bernard pays = 15% more

= 1.15 × $161

= $185.15

Therefore,

The amount more Bernard has to pay = $185.15 - $161

= $24.15 ≈ $24

Learn more about insurance premiums here:

brainly.com/question/3053945

3 0

2 years ago

Deshawn invests $5,000 in a savings account that earns 6% annual interest, compounded continuously. How long will it take to dou

koban [17]

Answer:

Step-by-step explanation:

Investment = $5,000

Annual Interest = 5%

5000/100 x 5/1

50×5 = 250

First year interest = $250

Therefore, 250 x 20

= $5,000

It will take him 40 years.

6 0

4 years ago

Help..................

Mashcka [7]

Im sorry, it is too blurry. I cannot read it.

4 0

3 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