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ipn [44]
3 years ago
13

A 15-foot flagpole leans slightly, such that it makes an 80° angle with the ground. The shadow of the flagpole is 10 feet long w

hen the sun has an unknown angle of elevation. How could the angle of elevation of the sun, x, be determined?
Mathematics
1 answer:
Naya [18.7K]3 years ago
5 0

Answer:   48°

<u>Step-by-step explanation:</u>

The shadow is the adjacent side and the length of the flag is the hypotenuse

cos\ \theta=\dfrac{adjacent}{hypotenuse}\\\\\\cos\ \theta=\dfrac{10}{15}\\\\\\cos^{-1}(cos\ \theta)=cos^{-1}\bigg(\dfrac{10}{15}\bigg)\\\\\\.\qquad \qquad \boxed{\theta=48^o}

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Which function represents the area with respect to time?
zhuklara [117]
It’s C I think but I am not sure about it
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3 years ago
- If MML = (11x - 5), mJK = (4x + 26)° and mZMNL = (10x – 12),<br><br><br><br> find mZMNL.
GenaCL600 [577]

Answer:

m∠ MNL = 78º

Step-by-step explanation:

<u>Arcs meassures </u>

mML = (11x - 5)º

mJK = (4x + 26)°

<u>Angle meassure</u>

m∠MNL = (10 x – 12)º

______________________

angle measure = average of arcs

m∠MNL  = (mML+ mJK)/2

Replacing

(10x – 12)º =  ((11x - 5)º + (4x + 26)° )/ 2

10x -12 =  (15x + 21 )/2

10x - 7.5x = 10.5 + 12

2.5x = 22.5

x= 9

m∠MNL = (10 x – 12)º

             = 90 - 12

             = 78º

3 0
3 years ago
The area of a square is 49 square meters. What is the length (in meters) of one side of the square?
AlladinOne [14]

Answer:

7

Step-by-step explanation:

8 0
3 years ago
1/4(x+2)–1/3(x–1)=1/2(x+1)​
Tanzania [10]
The answer is x=4/7 :)
7 0
3 years ago
Read 2 more answers
A 12 foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the
ira [324]

Using Pythagoras theorem, the top of the ladder moving down when the foot of the ladder is 3 feet from the wall is of -0.518 feet/sec.                      

Let distance from the wall to the foot of the ladder is 'x' feet and the height of the top of the ladder is 'y' feet.

Pythagoras theorem, x^{2} + y^{2} = (12)^{2}       --->(1)

Given,\frac{dx}{dt}= 2feet/second   at x=3

Put x=3 in Pythagoras theorem equation (1)

(3)^{2} + y^{2} = 144

         y^{2} = 144 - 9

        y^{2}  =  135

        y = 11.61

Derive equation (1) w.r.t to 't'

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0                ---->(2)

substitute the value of 'x', 'dx/dt' and 'y' in equation (2), we get the fast of the top of the ladder moving down when the foot of the ladder is 3 feet from the wall

2(3)(2) + 2 (11.61)\frac{dy}{dt}  = 0

12 + 23.22 \frac{dy}{dt}  = 0

                  \frac{dy}{dt}= \frac{-12}{23.22}

                  \frac{dy}{dt} = -0.518

Hence,  using Pythagoras theorem the top of the ladder moving down when the foot of the ladder is 3 feet from the wall is of -0.518 feet/sec.  

Learn more about Pythagoras theorem here

brainly.com/question/21511305  

#SPJ4    

     

5 0
2 years ago
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