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mamaluj [8]
3 years ago
5

Find the area of a trapezoid based on the information given below. H=4√2 b1=5√2 b2=3√2

Mathematics
1 answer:
alexgriva [62]3 years ago
3 0
The formula of the area of the trapezoid:
A=\dfrac{b_1+b_2}{2}\cdot h
We have:
b_1=5\sqrt2;\ b_2=3\sqrt2;\ h=4\sqrt2
Substitute:
A=\dfrac{5\sqrt2+3\sqrt2}{2}\cdot4\sqrt2=\dfrac{5\sqrt2}{2}\cdot4\sqrt2=\dfrac{5\cdot4\cdot\sqrt2\cdot\sqrt2}{2}\\\\=\dfrac{20\cdot2}{2}=20
Used: \sqrt{a}\cdot\sqrt{a}=a

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Write 6600 as a product of its prime factors. (show your working out)
dem82 [27]

Answer: 2 × 3 × 5 × 11 = 330.

Step-by-step explanation: Since, the prime factors of 6600 are 2, 3, 5, 11. Therefore, the product of prime factors = 2 × 3 × 5 × 11 = 330.

8 0
3 years ago
Which one is bigger 3.5km or 3,000 km
Alla [95]

Answer:

Converting kilometers to meters could never be simpler with our accurate length converter! ... Meters and kilometers both belong to the metric length scale, with these two units of measurement being on the larger side. ... 1 Kilometer equals 1000 Meters (1km = 1000m) ... 3 Kilometers equal 3000 Meters (3km = 3000m) ...

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
50
ozzi

Answer:

90 km, N 46° E

Step-by-step explanation:

<em>A jet flies due North for a distance of 50 km and then on a bearing of N 70° E for a further 60 km. Find the distance and bearing of the jet from its starting point.</em>

Look at the diagram I drew of this scenario. You can see the jet flies North for 50 km, and then turns at a 70° angle to fly another 60 km. We want to find the distance from the starting point, SP, to angle C (labeled).

This will be the jet's distance from its starting point.

In order to find the bearing of the jet from its starting point, we will need to find the angle formed between distances b and c, labeled angle A.

The <u>Law of Cosines</u> will allow us to use two known sides and one known angle to solve for the sides opposite of the known angle.

In this case, the known angle is 110° (angle B) so we will use the <u>Law of Cosines</u> respective to B.

  • b² = a² + c² - 2ac cosB

Substitute the known values into the equation and solve for b, the distance from the starting point (A) to the endpoint (C).

  • b² = (60)² + (50)² - 2(60)(50) cos(110°)
  • b² = 6100 -(-2052.12086)
  • b² = 8152.12086
  • b = 90.28909602
  • b ≈ 90 km

The distance of the jet from its starting point is 90 km. Now we can use this b value in order to calculate angle A, the bearing of the jet.

The <u>Law of Cosines</u> with respect to A:

  • a² = b² + c² - 2bc cosA

Substitute the known values into the equation and solve for A, the bearing from the starting point (clockwise of North).

  • (60)² = (90.28909602)² + (50)² - 2(90.28909602)(50) cosA
  • 3600 = 8152.12086 - 6528.909602 cosA
  • -4552.12086 = -6528.909602 cosA
  • 0.6972252853 = cosA
  • A = cos⁻¹(0.6972252853)
  • A = 45.79519
  • A ≈ 46°

The bearing of the jet from its starting point is N 46° E. This means that it is facing northeast at an angle of 46° clockwise from the North.

7 0
3 years ago
What is the area of this figure?<br><br><br> 49.5 in²<br><br> 66.5 in²<br><br> 84 in²<br><br> 91 in²
zubka84 [21]

The given figure can be divided in two parts 1) Square with side 7 in 2) A right triangle with base 12 in and height 7 in.

1) Area of square = side^{2}

Side of the square is 7 in.

Area = 7^{2} =49in^{2}

2) Area of triangle = \frac{1}{2} .base . height

Area of triangle =\frac{1}{2} (12)(7) =42in^{2}

Area of polygon= Area of square+ Area of triangle

Area of polygon = 49+42 =91 square in.

Area = 91in^{2}

3 0
3 years ago
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IgorLugansk [536]

Answer:

$945.08

Step-by-step explanation:

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