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4 years ago

When should the direct pattern be used to communicate negative nows?

1 answer:

3 0

If you are referring to news* instead of "nows", the direct pattern should be used when firmness is necessary and the receiver may overlook look it. This provides for a clear explanation and understanding.

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If you spin a spinner 2 times, what is the probability the sum of 2 numbers will be less than or equal to 9? (express your answe

kupik [55]

Answer: 2/8 i think

Step-by-step explanation:

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3 years ago

Cuanto es r2-2r-7=0

vekshin1

Answer:

Step-by-step explanation:

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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:

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