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3 years ago

Please tell me what to multiply by like the equations (finding the surface area)

Mathematics

2 answers:

Alborosie3 years ago

5 0

The top part of the shape would be (8×6)×1/2 to get that side. since it's the same as the bottom, multiply by that by two. so as of right now, you have two sides done (top and bottom sides). Next, lets do the side closest to us in the picture. since its a rectangle, all you have to do is 8×4. Add that to the answer we got earlier from the top and bottom sides. that gives us three sides done. we are going to do the one behind it next. That's a triangle as well. we will do 10×4 to get the area of that side. we will add that to get the total of 4 out of the 5 sides. Lastly, we will do the remaining side. 6×4. Add the answer of 6×4 to the earlier answer and that will get you the suface area of the whole shape. If you have follow up questions or wondering where i got each of the numbers from, comment and i will get back to you. Hope this helps and ignore any typos.

cupoosta [38]3 years ago

3 0

Length time weith time highth

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3 years ago

The area of a rectangular wall of a barn is 153 square feet. Its length is 8 feet longer than the width. Find the length and wid

r-ruslan [8.4K]

Answer and Step-by-step explanation:

Area (A) = 153

Length (l) = 8 + w

Width (w) = w

A = lw

153 = (8 + w)(w)

153 = $w^{2} + 8w$

0 = $w^{2} + 8w$ - 153

$w^{2} -9w + 17w - 153$

(w - 9)(w + 17)

w = 9

w = -17

We can't have a negative length, so we only use 9.

w equals 9.

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3 years ago

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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:

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

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:

$\boxed{t=4.03s}$

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4 years ago