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Ratling [72]
3 years ago
6

What is the degree of the monomial 15x^3y^9? (1 point)?

Mathematics
1 answer:
Contact [7]3 years ago
8 0
Pretty sure it's 12.
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If 15 baseballs weighs 75 ounces how many baseballs with 15 ounces
igor_vitrenko [27]

Answer:

5

Step-by-step explanation:

75/15=5

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Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficie
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Answer:

The differential equation will be like the one shown below

Step-by-step explanation:

Data:

Let the equation be given as:

y(4) + 8y' = 6

The equation will be expressed linearly as follows:

y(4) + 8\frac{dy}{dx}  = 6

8\frac{dy}{dx}+ 4y = 6

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I added a screeshot thank you
atroni [7]

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4 0
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I need help on this question plz
soldi70 [24.7K]

Cylinders are 3D figures, as they are solids with volume and take up space.

2D figures are flat figures that you can draw, like a square or circle.

1D figures are actually just a line, a segment, or a point, etc.

8 0
3 years ago
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.

5 0
3 years ago
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