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3 years ago

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Two rockets, A and B, are shot from two different launch pads. The path of rocket A can be represented by the quadratic function

A(t) = −(t − 8)2 + 535, where height, A(t), is in meters, and time, t, is in seconds. The path of rocket B is shown below.

1 answer:

oksano4ka [1.4K]3 years ago

4 0

<span>A(t) = −(t − 8)2 + 535
You would like to find the maximum of this function:</span> −(t − 8)^2<span>This part is always negative or zero as a number squared cannot be negative and you multiply by -1: Thus the maximum of this part MAX:</span>−(t − 8)^2=0
<span>The max will be when t=8 and its value is 535

</span>

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2√72 + 5√242 what is the answer for this question please can u tell me the right answer

Radda [10]

Answer:

67 $\sqrt{2}$

Step-by-step explanation:

Assuming you require to simplify the expression.

Using the rule of radicals

$\sqrt{a}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

Simplifying the radicals

$\sqrt{72}$ = $\sqrt{36(2)}$ = $\sqrt{36}$ × $\sqrt{2}$ = 6 $\sqrt{2}$

$\sqrt{242}$ = $\sqrt{121(2)}$ = $\sqrt{121}$ × $\sqrt{2}$ = 11 $\sqrt{2}$

Then

2 $\sqrt{72}$ + 5 $\sqrt{242}$

= 2(6 $\sqrt{2}$ ) + 5(11 $\sqrt{2}$ )

= 12 $\sqrt{2}$ + 55 $\sqrt{2}$

= 67 $\sqrt{2}$

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3 years ago

Find the 2nd Derivative:<br> f(x) = 3x⁴ + 2x² - 8x + 4

ad-work [718]

Answer:

$f''(x)=36x^2+4$

Step-by-step explanation:

Let's start by finding the first derivative of $f(x)= 3x^4+2x^2-8x+4$ . We can do so by using the power rule for derivatives.

The power rule states that:

$\frac{d}{dx} (x^n) = n \times x^n^-^1$

This means that if you are taking the derivative of a function with powers, you can bring the power down and multiply it with the coefficient, then reduce the power by 1.

Another rule that we need to note is that the derivative of a constant is 0.

Let's apply the power rule to the function f(x).

$\frac{d}{dx} (3x^4+2x^2-8x+4)$

Bring the exponent down and multiply it with the coefficient. Then, reduce the power by 1.

$\frac{d}{dx} (3x^4+2x^2-8x+4) = ((4)3x^4^-^1+(2)2x^2^-^1-(1)8x^1^-^1+(0)4)$

Simplify the equation.

$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x^1-8x^0+0)$
$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8(1)+0)$

$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8)$
$f'(x)=12x^3+4x-8$

Now, this is only the first derivative of the function f(x). Let's find the second derivative by applying the power rule once again, but this time to the first derivative, f'(x).

$\frac{d}{d} (f'x) = \frac{d}{dx} (12x^3+4x-8)$

$\frac{d}{dx} (12x^3+4x-8) = ((3)12x^3^-^1 + (1)4x^1^-^1 - (0)8)$

Simplify the equation.

$\frac{d}{dx} (12x^3+4x-8) = (36x^2 + 4x^0 - 0)$
$\frac{d}{dx} (12x^3+4x-8) = (36x^2 + 4(1) - 0)$
$\frac{d}{dx} (12x^3+4x-8) = (36x^2 + 4 )$

Therefore, this is the 2nd derivative of the function f(x).

We can say that: $f''(x)=36x^2+4$

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