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2 years ago

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Mathematics

1 answer:

andreyandreev [35.5K]2 years ago

4 0

B is the answer fam because I did this before and b was the answer

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Select the correct answer.

vovangra [49]

Step-by-step explanation:

let f(x) be y

∴ y = 10x/9 +11

(y - 11 )× 9/ 10 = x

Interchange x->y and y->x

(x - 11)×9/10 = y

here y is your f⁻¹(x)

∴ f⁻¹(x) = (x - 11 )× 9/10

or, f⁻¹(x) = 9x/10 - 99/10

Option (D) is correct

5 0

3 years ago

Read 2 more answers

The length of her scale drawing is 8 inches, and the width is 14 inches. The actual length of the MP3 player is 4 centimeters, a

Korvikt [17]

If the question is, "What is the scale factor." That would be a scale (factor) of 2.

6 0

4 years ago

A circular plate has a crack along the line AB, as shown below: A circular plate is shown. The center of the plate is labeled as

makkiz [27]

Well, since point A represents the center, and B represents a point on the outer line of the circle, segment AB would represent the radius, since the radius represents the length from the center to the outside of a circle

Hope this helps

8 0

4 years ago

Read 2 more answers

Determine the vertex of the function f(x)=3x^2-6x+13

Natasha_Volkova [10]

Answer:

Step-by-step explanation:

hello :

f(x)=3x^2-6x+13

1) a=3 and b= -6

2) the x coordinate of the vertex is : x= -b/2a so : x= -(-6)/6 = 1

3) the y coordinate by x= 1 is : f(1) = 3(1)²-6(1)+13 = 10

7 0

3 years ago

Jean throws a ball with an initial velocity of 64 feet per second from a height of 3 feet. Write an equation and answer the ques

Ganezh [65]

<h2>a. What is your equation?</h2>

This is a problem of projectile motion. A projectile is an object you throw with an initial velocity and whose trajectory is determined by the effect of gravitational acceleration. The general equation in this case is described as:

$h(t)=-\frac{1}{2}gt^2+v_{0}t+h_{0}$

Where:

$h(t): \ height \ at \ any \ time \\ \\ g: \ acceleration \ due \ to \ gravity \ 9.8m/s^2 \ or \ 32.16ft/s^2 \\ \\ v_{0}= \ Initial \ velocity$

So:

$v_{0}=64ft/s \\ \\ h_{0}=3ft$

Finally, the equation is:

$h(t)=-\frac{1}{2}(32.16)t^2+(64)t+3 \\ \\ \boxed{h(t)=-16.08t^2+64t+3}$

<h2>b. How long will it take the rocket to reach its maximum height?</h2>

The rocket will reach the maximum height at the vertex of the parabola described by the equation $h(t)=-16.08t^2+64t+3$ . Therefore, our goal is to find $t$ at this point. In math, a parabola is described by the quadratic function:

$f(x)=ax^2+bx+c$

So the x-coordinate of the vertex can be calculated as:

$x=-\frac{b}{2a}$

From our equation:

$a=-16.08 \\ \\ b=64 \\ \\ c=3$

So:

$t=-\frac{64}{2(-16.08)} \\ \\ \boxed{t=1.99s}$

So the rocket will take its maximum value after 1.99 seconds.

<h2>c. What is the maximum height the rocket will reach?</h2>

From the previous solution, we know that after 1.99 seconds, the rocket will reach its maximum, so it is obvious that the maximum height is given by $h(1.99)$ . Thus, we can find this as follows:

$H_{max}=h(1.99)=-16.08(1.99)^2+64(1.99)+3 \\ \\ \boxed{H_{max}=66.68ft}$

So the maximum height the rocket will reach is 66.68ft

<h2>d. How long is the rocket in the air?</h2>

The rocket is in the air until it hits the ground. This can be found setting $h(t)=0$ , so:

$0=-16.08t^2+64t+3 \\ \\ Applying \ quadratic \ formula: \\ \\ t_{12}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ a=-16.08 \\ \\ b=64 \\ \\ c=3 \\ \\ t_{12}=\frac{-64 \pm \sqrt{64^2-4(-16.08)(3)}}{2(-16.08)} \\ \\ t_{1}=4.0264 \\ \\ t_{2}=-0.046$

We can't have negative value of time, so the only correct option is $t_{1}=4.0264$ and rounding to the nearest hundredth we have definitively:

$\boxed{t=4.03s}$

3 0

4 years ago