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2 years ago

Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.

Mathematics

1 answer:

Hatshy [7]2 years ago

3 0

Answer:

The change in the area of the rectangle is 0 square feet

Step-by-step explanation:

To solve this problem we have to calculate the area of the two rectangles that give us

first we have to know the formula to calculate the area of a rectangle

a = area

l = length = 12ft

w = width = 5ft

a = l * w

we replace the known walues

a = 12ft * 5ft

a = 60ft²

we do the same with the other rectangle, but first we have to calculate its sides

w = 5ft - (5ft * 20/100)

w = 5ft - 1ft

w = 4ft

l = 12ft + (12ft * 25/100)

l = 12ft + 3ft

l = 15ft

a2 = 15ft * 4ft

a2 = 60ft²

to calculate the change in the area we subtract (a - a2)

a - a2 =

60ft² - 60ft² = 0ft²

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Answer:

5/ 19 ≈ 0.26 = 26%

Step-by-step explanation:

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2 years ago

Three security cameras were mounted at the corners of a triangles parking lot. Camera 1 was 110 ft from camera 2, which was 137

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Answer:

Camera 2nd has to cover the maximum angle, i.e. $78.70^\circ$ .

Step-by-step explanation:

Please have a look at the triangular park represented as a triangle $\triangle ABC$ with sides

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b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles $\angle A, \angle B, \angle C$

As per Law of cosine:

$cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}$

Putting the values of a,b and c to find out angles $\angle A, \angle B, \angle C$ .

$cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ$

$cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ$

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Camera 2nd has to cover the maximum angle, i.e. $78.70^\circ$ .

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