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2 years ago

If each unit on the grid represents 2.5 inches, find the height of the triangle.

Mathematics

2 answers:

MaRussiya [10]2 years ago

7 0

Answer:

Step-by-step explanation:

on edge

pentagon [3]2 years ago

3 0

The answer is 15 inches

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50 POINTS I WILL GIVE BRAINLIEST!!! HELPPPPPPPP....explain how to construct congruent line segments..

tigry1 [53]

Answer:

1. Construct a segment XY on the a sketch.

2. Elsewhere on the sketch, draw a line and create a point on the line . Label the point P.

3. Select point P and line segment XY. Construct circle by center and radius.

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5. Select point P and point Q. Construct segment PQ.

6. Select the circle, line, and other point of intersection. Hide objects. Only segment XY and segment PQ should be visible on the sketch.

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2 years ago

X+4/8 = 1/2

Nina [5.8K]

Answer:

x=0

Step-by-step explanation:

4/8=1/2, so:

x+1/2=1/2

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3 years ago

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A square has an area that is less than 100 m. What is a reasonable range for the graph of the

WITCHER [35]

1-9 I’m pretty sure but idk

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3 years ago

By treating savings as an ___ it guarantees that you will save the money instead of spending it

Mekhanik [1.2K]

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Step-by-step explanation:

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3 years ago

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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:

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

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

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

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

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

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

6 0

3 years ago