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tekilochka [14]
3 years ago
12

Please show me how to do this!!

Mathematics
1 answer:
Murrr4er [49]3 years ago
7 0
(2x^4 + 3x^2)(x^2-1)
you multiply and you'll get:
2x^8-2x^4+3x^4-3x^2
then add the coefficients with x^4
= 2x^8+1x^4-3x^2

 the coefficient for x^4 is 1, so the answer is B
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A brick is thrown upward from the top of a building at an angle of 250 above the horizontal, and with an initial speed of 15 m/s
mojhsa [17]

Answer:

(a) 25.08 m

(b) 2.05 m

Step-by-step explanation:

Let the height of the building be 'h'.

Given:

Angle of projection is, \theta=25°

Initial speed is, u=15\ m/s

Time of flight is, t=3.0\ s

(a)

Consider the vertical motion of the brick.

Vertical component of initial velocity is given as:

u_y=u\sin\theta\\\\u_y=15\sin(25)=6.34\ m/s

Vertical displacement of the brick is equal to the height of the building.

So, vertical displacement = -h (Negative sign implies downward motion)

Acceleration is due to gravity in the downward direction. So,

Acceleration is, g=-9.8\ m/s^2

Now, using the following equation of motion;

-h=u_yt+\frac{1}{2}gt^2\\-h=6.34(3)-\frac{9.8}{2}(3)^2\\\\-h=19.02-44.1\\\\-h=-25.08\\\\h=25.08\ m

Therefore, the building is 25.08 m tall.

(b)

Let the maximum height be 'H'.

At maximum height, the vertical component of velocity is 0 as the brick stops temporarily in the vertical direction.

So, v_y=0\ m/s

Now, using the following equation of motion, we have:

v_y^2=u_y^2+2gH\\\\0=(6.34)^2-2\times 9.8\times H\\\\19.6H=40.2\\\\H=\frac{40.2}{19.6}=2.05\ m

Therefore, the maximum height of the brick is 2.05 m.

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2. Subtract 5(7z=-13)

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Answer:

the correct answer would be 2.25 or 2

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