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3 years ago

Factorise 2x squared +6x

1 answer:

7 0

Hey there.

When we are factoring binomials, it's significant that we ensure that we simplify our terms in the end, so let's work our way there.

2x^2 + 6x.

First, let's look for any terms that we have in common. As both of our coefficients are even numbers and have a variable, we can divide them by at least 2x. Divide both terms by 2x.

2x(x + 3) is now our answer, and is our fully factored answer.

I hope this helps.

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The value 9 is 200% of what number?

kykrilka [37]

Answer:

Step-by-step explanation:

If 100% is 9 then you need to multiply by 2 to get 200% which is 18.

3 0

3 years ago

Read 2 more answers

Solve the equation p^2 + 4p = 1 by completing the square.

igor_vitrenko [27]

Answer:

p = -2 ±sqrt( 5)

Step-by-step explanation:

p^2 + 4p = 1

Take the coefficient of p

Divide by 2

4/2 =2

Square it

2^2 = 4

Add it to each side

p^2 + 4p+4 = 1+4

(p+2) ^2 = 5

Take the square root of each side

sqrt((p+2) ^2) =±sqrt( 5)

p+2 = ±sqrt( 5)

Subtract 2 from each side

p+2-2 = -2 ±sqrt( 5)

p = -2 ±sqrt( 5)

7 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 (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 $t = 6$ .)