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shtirl [24]
3 years ago
5

The sum of 3 consecutive even integers is 78. What is the largest of the integers?

Mathematics
2 answers:
Vsevolod [243]3 years ago
8 0

Answer:

28

Step-by-step explanation:

Let the first even integer be a.

The second even integer = a + 2

The third even integer = 2 + (a + 2) = a + 4

The sum of three consecutive even integers = 78

78 = a + (a + 2) + (a + 4)

78 = 3a + 6

72 = 3a

a = 24

The largest even integers is the third even integer = a + 4 = 24 + 4 = 28

Maksim231197 [3]3 years ago
3 0

Answer:

is the sum 78 or 7,878 (i.e. was there a stutter)?

In any event, this sum must be even, so 3 consecutive even integers would be n, n+2 & n+4 whose sum is 3n + 6.

Set that to either 78 or 7878, whichever is intended, & solve for n.   If 78 is their grand total, then those 3 consecutive even numbers have to be 24, 26 & 28.

No stress!  Just logic! Keep the faith!

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Let the height of the building be 'h'.

Given:

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Time of flight is, t=3.0\ s

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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,

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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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3 years ago
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Answer:

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Step-by-step explanation:

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