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3 years ago

Point A is located at (1, 5), and point A′ is located at (-2, 1). What is the distance from B to B′ ?

Mathematics

1 answer:

Lunna [17]3 years ago

5 0

Answer:

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Step-by-step explanation:

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Natasha2012 [34]

Answer:

Step-by-step explanation:

10 x 8 equals the area for a square/rectangle then multiplied by .5 equals area for a triangle. You multiply because triangles are half of a square.

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3 years ago

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3b×+7c 2×2+4 3×3_6×+5 7y +8s_3

PilotLPTM [1.2K]

Answer:

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4 years ago

A company logo is shaped like an equilateral triangle with 2-in.-long sides. What is the height of the logo? Round to the neares

deff fn [24]

Answer:

B. 1.7 in

Step-by-step explanation:

Divide the triangle into 2 congruent right angle triangles

With hypotenuse: 2

Base: 2/2 = 1

Using pythagoras theorem

2² = 1² + h²

h² = 3

h = sqrt(3)

h = 1.732050808

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3 years ago

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Ann [662]

Answer:

X = 3x + 3

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3 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

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