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maks197457 [2]
3 years ago
7

(1.2)Consider the diagram below. Identify a point.

Mathematics
1 answer:
Fynjy0 [20]3 years ago
6 0

Option B is correct. <em>Point F</em><em> is the </em><em>only point </em><em>according to the option given</em>

A point is defined as an exact location with no size, length, or even width.

From the diagram shown, we have points, plane, and line segments.

  • The only plane that we have is plane R making option C incorrect.

  • A line is defined as a straight figure connected by two points. From the diagram, the line segments that we have are AN, BN, and AB falsifying options A and D.

The points are C, D, E, and F. Hence the correct option is B.<em> Point F</em><em> is the </em><em>only point </em><em>according to the option given</em>

Learn more here: brainly.com/question/13099718

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TW +DW

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iVinArrow [24]
The measure of an angle formed by two tangent lines intersecting at a point outside the circle is equal to one-half their intercepted arcs.
XYZ = 1/2(XSZ - XZ)
XYZ = 1/2 (220 - 140)
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The measure of angle XYZ is 40 degrees.

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6 0
4 years ago
The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
2 years ago
72 divided by 2 times 4 divided by 3
vovangra [49]
72 ÷ 2= 36

36 × 4= 144

144 ÷ 3= 48
7 0
3 years ago
Read 2 more answers
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