A balloon rises to the ceiling of a gymnasium. You want to find the distance from
the ground to the balloon. You use a cardboard square to line up the balloon and
the ground. Your friend measures the vertical distance from the ground to your
eye and the distance from you to the gym wall. Approximate the distance from
the ground to the balloon.

Solution given;

∆ABD & ∆ABC & ∆ACD are right angled triangle.

AC=8ft

BC=4ft.

BD=?

In ∆ ABC

Tan A=P/b

<BAC=Tan-¹ (4/8)

=26.565°

again

<CAD=90°-26.564=63.436°

In ∆ ACD

Tan <CAD=P/b

Tan (63.436°)=CD/8

CD=2×8=16

So

BD=4+16=20ft.

Therefore Approximate the distance from

Approximate the distance fromthe ground to the balloon is 20ft.

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.

she will most likely have 33  

step-by-step explanation: