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aivan3 [116]
2 years ago
11

A 20 ft. ladder is used against a 15 ft. wall. What is the measure of the angle made by the ladder and the ground (nearest whole

degree)? How far is the ladder from the wall on the ground (to the nearest tenth)?
Mathematics
1 answer:
Brrunno [24]2 years ago
6 0

Step-by-step explanation:

Given that,

The length of a ladder, H = 20 feet

The height of the wall, h = 15 ft

We know that,

\sin\theta=\dfrac{h}{H}

h is perpendicular and H is hypotenuse

So,

\sin\theta=\dfrac{15}{20}\\\\\theta=\sin^{-1}(\dfrac{15}{20})\\\\\theta=48.59^{\circ}

Now using Pythagoras theoerm,

b=\sqrt{H^2-h^2}\\\\b=\sqrt{20^2-15^2}\\\\b=13.2\ ft

Hence, the angle made by the ladder and the ground is 48.59° and the ladder is 13.2 feet from the wall on the ground.

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S_A_V [24]
<h3>The distance covered by Drew in long jump is either 7 ft 5 inches OR 7 ft  6 inches.</h3>

Step-by-step explanation:

Here, the needed table is attached for the reference.

Now, as we can see from the table:

The distance covered by  Cindy  = 2 yards, 1 foot 3 inches

The distance covered by  Tyrette   =  7 feet, 2 inches

The distance covered by Nina   =   2 yards, 1 foot, 1 inch

The distance covered by  Monique   = 7 feet, 4 inches

As we know : 1 yard  =  3 feet

So, the distance covered  by :

Cindy  =  2 yards, 1 foot 3 inches = 3 ft x (2)  + 1 ft +  3 in  = 7 ft 3 in

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Nina   =   2 yards, 1 foot, 1 inch  = = 3 ft x (2)  + 1 ft +  1 in  = 7 ft 1 in

Monique   = 7 feet, 4 inches

So by comparing all distances, we can see that:

The maximum distances jumped by all four  is 7 ft 4 inches.

The distance covered by Drew is less than 7 ft 7 in.

So, he must have jumped 7 ft 5 inches OR 7 ft  6 inches.

Hence, the distance covered by Drew in long jump is either 7 ft 5 inches OR 7 ft  6 inches.

4 0
3 years ago
How is the pythogorean theorem used to determine the distance formula
Fiesta28 [93]

Answer:

The distance formula uses the coordinates of points and the Pythagorean Theorem to calculate the distance between points. If A and B form the hypotenuse of a right triangle, then the length of AB can be found using this formula, leg2 + leg2 = hypotenuse2 (or you can use A squared + B squared = C squared)

Am not a real genius but I hope that answers your question! ^w^

5 0
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Piont P is plotted on the coordinate grid. If point S is 10 units to the left of point P, what are the coordinates of point S?
frozen [14]

Answer:

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Step-by-step explanation:

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7 0
2 years ago
The area of a rectangle invitation card is 3/10 square foot. Given that the length of the card is 4/5 foot, find its perimeter.
Leokris [45]
Area is length times width [A=(L)(W)], if we know that the area of the card is 3/10 and the length is 4/5 we can set up the equation 3/10=(4/5)(W) where W is width. Solve for W by deciding both sides by 4/5 and we get that W=3/8. To find the perimeter of the card were have to add up all the sides of the card so 3/8+3/8+4/5+4/5 which equals 47/20
8 0
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If $206.40 amounts to<br>$237.36 in 2 years, find the<br>rate of simple interest per<br>annum​
saveliy_v [14]

Answer:

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Step-by-step explanation:

You get the difference from the new number and the original. In this case, it is 30.96.  I then divided the original by  30.96. The answer turned out to be .15. I doubled checked my answer by timing the original by .15 and adding it. But, I had to divide .15 by 2 so I could get the amount per year. So, it would be 7.5% per year of interest.

8 0
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