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3 years ago

If u answer ill give brainlist

Mathematics

1 answer:

makkiz [27]3 years ago

3 0

Answer:

120; 643 units squared

Step-by-step explanation:

Perimeter: Add all the sides, so 120 units

_________________________________

Area: First multiply 19 by 24 for the first shape within this other shape, so That equals 456 units squared. Since we know that one whole side is 36 we can subtract 19 from it to get the lower side of the smaller shape and then multiply that by 11. So the smaller side ends up being 17. So 17 times 11 equals: 187 units squared. Then just add the areas together 187+456=643 units squared.

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Answer:

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Step-by-step explanation:

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3 years ago

True or false: the following equation is in standard form. (x - 4.2)^2 - 3 = 0

natita [175]

Answer:

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Step-by-step explanation:

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3 years ago

PLEASE HELP FAST!!!!!!!!

seropon [69]

Answer:

Option B. -j = -h/-k is not correct

Step-by-step explanation:

As from the given scenario both the negative signs will be cancelled out giving positive j : -h/-k = -j

First option has: -j = -h/k In this case also the negative sign from both sides would be cancelled out.

Second option has: -h/-k = -j In this case negative signs cannot be cancelled out.

Third option has: h/-k = -j , negative sign would be cancelled from both sides.

Fourth option has: h/k = j , no negative sign on either side.

i hope it will help you!

7 0

3 years ago

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Furkat [3]

Answer:

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2n + 20

4 0

3 years ago

Read 2 more answers

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

Nata [24]

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

a = 110 ft

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 $\angle A, \angle B, \angle C$ .

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

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

$cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ$

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

6 0

4 years ago