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Sedaia [141]
2 years ago
5

Suppose you are standing such that a 36-foot tree is directly between you and the sun. If you are standing 50 feet away from the

tree and the tree casts a 60-foot shadow, how tall could you be and still be completely in the shadow of the tree?

Mathematics
1 answer:
musickatia [10]2 years ago
7 0

Answer:

6 feet

Step-by-step explanation:

Given,

The height of tree, say AB ( A is top, B is bottom ) = 36 foot,

Length of its shadow, BE = 60 feet,

Suppose CD represents me, ( where C is top, D is bottom )

We have, BD = 50 feet,

∵ DE = BE - BD = 60 - 50 = 10 feet,

Using trigonometric ratio,

\tan A = \frac{CD}{DE}=\frac{AB}{BE}

\implies \frac{CD}{10}=\frac{36}{60}

CD = \frac{36}{60}\times 10 =\frac{36}{6}=6

Hence, I would be 6 feet tall.

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Family Video charges $10 for
Irina18 [472]

Answer:

4 movies, $20

Step-by-step explanation:

Given data

Family Video

charges $10 for a monthly membership and

$2.50 per movie

let the number of movies be x

and the charge for x movies be y

y= 10+2.5x----------1

Babe's  Movies

charges a $12

membership and $2 per

movie.

let the number of movies be x

and the charge for x movies be y

y= 12+2x----------2

equate 1 and 2

10+2.5x= 12+2x

collect like terms

12-10= 2.5x-2x

2= 0.5x

x= 2/0.5

x= 4 movies

put x= 4 in equation 1 or 2

y= 12+2x----------2

y= 12+2*4

y= 12+8

y= $20

4 0
3 years ago
7. Triangle FIP was dilated by a scale factor of centered at the origin to
Assoli18 [71]

Answer:

AB is parallel to A'B'.

DO,1/2 (1/2x, 1/2y) =

The distance from A' to the origin is half the distance from A to the origin.

Step-by-step explanation:

6 0
2 years ago
A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.50 kg mass in each outstre
Vlad1618 [11]

Answer:r'=0.327 m

Step-by-step explanation:

Given

N=2.85 rev/s

angular velocity \omega =2\pi N=17.90 rad/s

mass of objects m=1.5 kg

distance of objects from stool r_1=0.789 m

Combined moment of inertia of stool and student =5.53 kg.m^2

Now student pull off his hands so as to increase its speed to 3.60 rev/s

\omega _2=2\pi N_2

\omega _2=2\pi 3.6=22.62 rad/s

Initial moment of inertia of two masses I_0=2mr_^2

I_0=2\times 1.5\times (0.789)^2=1.867

After Pulling off hands so that r' is the distance of masses from stool

I_0'=2\times 1.5\times (r')^2

Conserving angular momentum

I_1\omega =I_2\omega _2

(5.53+1.867)\cdot 17.90=(5.53+I_o')\cdot 22.62

I_0'=1.397\times 0.791

I_0'=5.851

5.53+2\times 1.5\times (r')^2=5.851

2\times 1.5\times (r')^2=0.321

r'^2=0.107009

r'=0.327 m

7 0
3 years ago
Pls help I’m failing math I’ll brainlest
Keith_Richards [23]

Answer:

z=-9 - 5.4

hope it helps

Step-by-step explanation:

please mark me brainliest

4 0
2 years ago
Next Activity
astra-53 [7]
700%10=70

70•25=1,750 Calories

Brainliest would be cool, hope this helped
7 0
3 years ago
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