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katovenus [111]
3 years ago
6

A = 1/2h(B+b); A=77,B=10, b=1 I need help please and thank you

Mathematics
1 answer:
Inga [223]3 years ago
7 0

Answer:

H = 14

Step-by-step explanation:

Plug in the numbers.

77 = 1/2h (10+1)

77 = 1/2h (11)

7 = 1/2h

multiply by 2 to get rid of fraction

14 = h

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Find the value of each measure.
IrinaK [193]

Answer:

39

Step-by-step explanation:

4x+7=9x-33

4x-9x=-33-7

-5x=-40

x=8

4×8+7=39

4 0
3 years ago
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Base: 5 ft<br> area: 5 5/6 <br> what's the height( pls hurry&lt;3)
muminat
Answer:

The height is 1.666 or about 2.

Explanation:

Since area= b x h, you just divide the area by the base to get your answer (about 2).

Hope this helped!
3 0
3 years ago
One number is ten times another number. Their sum is 253.
s344n2d4d5 [400]

Answer:

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Step-by-step explanation:

230 is ten times larger than 23, and if you add them together is it 253

5 0
3 years ago
An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object
Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2  + \frac{1}{2} at^2\\3) v = v_0 + at

What do we already know?

  • v_0 = 29.4 ms^-1
  • The launch is straight up
  • a = -9.81 ms^-2 this is the gravitational acceleration g
  • s_0 = 0 m, since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, v = 0 ms^-2

v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2

this is the equation of the whole trajectory that object makes.

and if you solve this by making s = 0, you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

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Answer:

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