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VladimirAG [237]
2 years ago
5

What’s the area for each figure?

Mathematics
1 answer:
Vanyuwa [196]2 years ago
5 0

9514 1404 393

Answer:

  1. 52.5 in²
  2. 52.5 in²
  3. 76.97 in²
  4. total: 181.97 in²

Step-by-step explanation:

1, 2. The area of each triangle is given by the formula ...

  A = (1/2)bh . . . . . for base b and height h

The area of the given triangle is ...

  A = (1/2)(7 in)(15 in) = 105/2 in² = 52.5 in²

Triangles 1 and 2 have the same dimensions, so have the same area.

  Figure 1 area: 52.5 in²

  Figure 2 area: 52.5 in²

__

3. The area of the semicircle is given by the formula ...

  A = (1/2)πr² . . . . for a semicircle of radius r

This semicircle has a radius of 7 in, so an area of ...

  A = (1/2)(3.1416)(7 in)² ≈ 76.97 in²

  Figure 3 area: 76.97 in²

__

The total area is the sum of these:

  52.5 in² + 52.5 in² +76.97 in² = 181.97 in²

  Composite Figure area: 181.97 in²

_____

<em>Comment on decimal digits</em>

We cannot tell your rounding requirements. Here, we have rounded to the nearest hundredth. If you use the value 3.14 for pi, then your areas for the semicircle and composite figure will end in .93, not .97.

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Free_Kalibri [48]
<h3>Explanation:</h3>

1. PQ║TS, PQ ≅ TS, PT and QS are transversals to the parallel lines . . . given

2. ∠P ≅ ∠T . . . alternate interior angles at PT

3. ∠Q ≅ ∠S . . . alternate interior angles at QS

4. ΔPQR ≅ ΔTSR . . . ASA postulate

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You can use any pair of angles together with the sides PQ and TS. If you use the vertical angles and one of ∠T or ∠S, then you must invoke the AAS postulate for congruence, as the side is not between the two angles.

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3 years ago
4.<br> Which point on the number line<br> represents 0.3?
Dafna11 [192]

Answer:

Point C

Step-by-step explanation:

because the number is 0.3 you need to start at zero on the number line and go forward 3 marks. That will lead you to your answer of 0.3

5 0
2 years ago
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Given: cos θ=-4/5, sin x = -12/13, θ is in the third quadrant, 
USPshnik [31]

By definition of tangent,

tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)

Recall the double angle identities:

sin(2<em>θ</em>) = 2 sin(<em>θ</em>) cos(<em>θ</em>)

cos(2<em>θ</em>) = cos²(<em>θ</em>) - sin²(<em>θ</em>) = 2 cos²(<em>θ</em>) - 1

where the latter equality follows from the Pythagorean identity, cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1. From this identity we can solve for the unknown value of sin(<em>θ</em>):

sin(<em>θ</em>) = ± √(1 - cos²(<em>θ</em>))

and the sign of sin(<em>θ</em>) is determined by the quadrant in which the angle terminates.

<em />

We're given that <em>θ</em> belongs to the third quadrant, for which both sin(<em>θ</em>) and cos(<em>θ</em>) are negative. So if cos(<em>θ</em>) = -4/5, we get

sin(<em>θ</em>) = - √(1 - (-4/5)²) = -3/5

Then

tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)

tan(2<em>θ</em>) = (2 sin(<em>θ</em>) cos(<em>θ</em>)) / (2 cos²(<em>θ</em>) - 1)

tan(2<em>θ</em>) = (2 (-3/5) (-4/5)) / (2 (-4/5)² - 1)

tan(2<em>θ</em>) = 24/7

4 0
3 years ago
HELP PLEASE!!!!!BRAINLIEST
Greeley [361]
That is false because you have to subtract the exponents not divide
4 0
3 years ago
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PLZ IN NEED OF HELP
Serga [27]
The roots would be x=-5 and x=-4 but with that website you just do -5 and -4
6 0
3 years ago
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