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vagabundo [1.1K]
3 years ago
6

Question 18

Mathematics
1 answer:
ira [324]3 years ago
6 0

Answer:

The function which represents this sequence will be:

a_n=-15n+110

Hence, option (A) is true.

Step-by-step explanation:

Given the sequence

95, 80, 65, 50, ...

An arithmetic sequence has a constant difference 'd' and is defined by  

a_n=a_1+\left(n-1\right)d

computing the differences of all the adjacent terms

80-95=-15,\:\quad \:65-80=-15,\:\quad \:50-65=-15

As the difference is the same, so

d = -15

as

a_1=95

Thus, substituting d = -15, a_1=95 in the nth term of an arithmetic sequence

a_n=a_1+\left(n-1\right)d

a_n=-15\left(n-1\right)+95

a_n=-15n+110

Therefore, the function which represents this sequence will be:

a_n=-15n+110

Hence, option (A) is true.

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Step-by-step explanation:

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Simplify. 5y(4x-3z)
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Answer:

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NEED HELP ASAP
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4) -8^4

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7) -100

8) 30

There ya go! Pretty sure they're all right!



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3 years ago
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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.

5 0
3 years ago
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