a model rocket is launched into the air from ground level. the height, in feet, in feet, is modeled by p(x)=16x^2+32x, where x is the number of elapsed seconds. what is the total number of seconds the model rocket will be in the air?

Answers

Answer 1

Answer: Since you are given the function for the height of a rocket, we are to look for the value of X that gives us a value of p(x) = 0, which means that the model rocket has landed back onto the ground. One easy way to do this is by table method, meaning we substitute values of x to see if we can obtain the needed value of P(x). We should start from 1, since using 0 as our first value means that the rocket has not even launched up yet. Let us do this below:

x | p(x) = -16x^2+32x
--------------------------
1 | p(1) = -16(1)^2 + 32(1) = 16
2 | p(2) = -16(2)^2 + 32(2) = 0

Having obtained the desired result at x = 2, this means that the model rocket has been in the air for 2 seconds.

Answer 2

Answer:

Final answer:

The model rocket is in the air for 0 seconds according to the given equation, which implies an immediate impact upon launch.

Explanation:

In this mathematical scenario, the model rocket's height p(x) over time x (elapsed seconds) is given by the quadratic equation p(x) = 16x^2 + 32x. The total amount of time the rocket is in the air will be the point when the rocket returns to ground level. This occurs when p(x) = 0, which represents the rocket's height being zero feet above the ground.

We can find out when this occurs by solving the quadratic equation for x. We can rearrange the quadratic equation to 16x^2 + 32x = 0. Factoring out 16x gives us 16x(x + 2) = 0. Solving for x will give two potential solutions: x = 0 (the initial launch point) and x = -2. However, since time cannot be negative in this context, we discard the -2 and our answer is x=0 s, the total time the model rocket will be in the air after being launched is 0 seconds.

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What is the indefinite integral of 6sin(3t) dt?

Answers

Answer:

$\displaystyle \int {6sin(3t)} \, dt = -2cos(3t) + C$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals
[Indefinite Integrals] Integration Constant C

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int {6sin(3t)} \, dt$

Step 2: Integrate Pt. 1

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {6sin(3t)} \, dt = 6\int {sin(3t)} \, dt$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set u: $\displaystyle u = 3t$
[u] Differentiate [Basic Power Rule, Multiplied Constant]: $\displaystyle du = 3 \ dt$

Step 4: integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {6sin(3t)} \, dt = 2\int {3sin(3t)} \, dt$
[Integral] U-Substitution: $\displaystyle \int {6sin(3t)} \, dt = 2\int {sin(u)} \, du$
[Integral] Trigonometric Integration: $\displaystyle \int {6sin(3t)} \, dt = -2cos(u) + C$
Back-Substitute: $\displaystyle \int {6sin(3t)} \, dt = -2cos(3t) + C$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

$48 marked down by 20%

Answers

48-20%=48*(1-0.2)=48*.8= 38.4
$38.40

A force of 3 pounds acts in the direction of 42 degrees to the horizontal. The force moves an object along a straight line from the point (3,7) to the point (8,8), with distance measured in feet. Find the work done by the force in foot-pounds.

Answers

Answer:

11.37 lb-ft

Step-by-step explanation:

We are given that

Force,F=3 pounds

$\theta=42^(\circ)$

Let Point A (3,7) and point B(8,8)

We have to find the work done by the force.

$d=<(8-3)+(8-7)>=<5,1>$

$r=\mid d\mid=√(x^2+y^2)=√(5^2+1^2)=√(26)$ feet

We know that Work done by the force

$W=Frcos\theta$

Substitute the values

$W=3* √(26)cos42=11.37 lb-ft$

Hence, the work done by the force =11.37 lb-ft

Final answer:

The work done by a force of 3 pounds moving an object along a straight line from point (3,7) to (8,8) at an angle of 42 degrees with the horizontal is calculated to be approximately 11.7 foot-pounds.

Explanation:

The student's question relates to the concept of work done by a force. According to the equation for work, W = Fd cos θ, the work done by a force is the product of the magnitude of the force, the distance over which it acts and the cosine of the angle between the force and the displacement vectors. In this case, the force is 3 pounds, the displacement can be calculated using Pythagorean theorem from the points (3,7) to (8,8) yielding approximately 5.1 feet, and the angle with the horizontal is 42 degrees. Using these values, the work done can be calculated as:

W = 3 pounds * 5.1 feet * cos(42 degrees) = approximately 11.7 foot-pounds

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Identify each quantity as a parameter or statistic Write "parameter" or "statistic" (without quotations). a) X: parameter b) S: statistic c) o: statistic هما d) u: parameter

Answers

The classification of each quantity as either a parameteror statistic is given thus :

X: parameter
S: statistic
o: statistic
u: parameter

a) X: parameter - A numerical summary of a population, and X represents a population parameter.

b) S: statistic - A numerical summary of a sample, and S represents a sample statistic.

c) o: statistic - Similarly, "o" represents a sample statistic, as it is used to denote the sample standard deviation.

d) u: parameter - "u" represents a population parameter, often used to denote the population mean.

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Answer:

Step-by-step explanation:

Length of a rectangle is 2.5 cm and it’sarea is 12.5cm2. Ram found that the breadth of the rectangle is 5 cm while Rahim found that the breadth is 10 cm. Who is correct? Justify your answer.

Answers

Step-by-step explanation:

ram is true because when 2.5 is multiplied to .5 centimeters then answered becomes 12.5 centimeters square

For wich values of x does 2x^2+10x

Answers

2x^2 + 10x doesn't do anything, unless it's equal to something.

All you've got so far is a number. We can't even tell what it is
until we know what number 'x' is.

Is there another little part of the question that you didn't copy ?

Factor out at 2 so 2(x+5)(x+0)
Set them equal to 0 so
2=0 that's not a solution
X+5=0 so x=-5 that works
And the last one x=-0 well negative 0 doesn't exist so the only value that works is -5

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