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rjkz [21]
3 years ago
10

An object is launched with a velocity at 64 feet per second from a platform 30 feet high. the function H(t)= -16t2+64t+30 repres

ents the function where T is time in seconds, and H is height in feet.
what is the maximum height of the object ever reaches?
a) 92 ft
b) 94 ft
c) 2 ft
d) 94 ft

when does it reach the maximum height?

a) 2 sec
b) 94 sec
c) -2 sec
d) 3 sec
Mathematics
1 answer:
PolarNik [594]3 years ago
4 0
To find the maximum or minimum value of a function, we can find the derivative of the function, set it equal to 0, and solve for the critical points.

H'(t) = -32t + 64

Now find the critical numbers:

-32t + 64 = 0

-32t = -64

t = 2 seconds

Since H(t) has a negative leading coefficient, we know that it opens downward. This means that the critical point is a maximum value rather than a minimum. If we weren't sure, we could check by plugging in a value for t slightly less and slighter greater than t=2 into H'(t):

H'(1) = 32

H'(3) = -32

As you can see, the rate of change of the object's height goes from increasing to decreasing, meaning the critical point at t=2 is a maximum.

To find the height, plug t=2 into H(t):

H(2) = -16(2)^2 +64(2) + 30 = 94

The answer is 94 ft at 2 sec.
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Answer: The answer is one will have 8 and one will have 9 so roughly 8

Step-by-step explanation

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2 years ago
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Answer: 22 minutes

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8 0
3 years ago
3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k
Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt

Evaluate the integral to solve for y :

\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt

\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x

\displaystyle y = x^3+2x^2+kx - 2

Use the other known value, f(2) = 18, to solve for k :

18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}

Then the curve C has equation

\boxed{y = x^3 + 2x^2 + 2x - 2}

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2

The slope of the given tangent line y=x-2 is 1. Solve for a :

3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1

So, the point of contact between the tangent line and C is (-1, -3).

7 0
2 years ago
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ch4aika [34]

Answer:

A. (4x + 9) (4x − 9) is the correct answer.


6 0
3 years ago
3+2/3(3-x)<-7 please help I am stuck
asambeis [7]
Solve the problem then state if it's true or false- this is an order of operation problem. 3+2/3(3-X) this is the problem. Now distribute, 3+ 2/3(3)- 2/3(x)= 3+2- 2/3x so the final expression would be 5-2/3x I don't know about the <-7 part but that's the solution to the first one
8 0
3 years ago
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