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zloy xaker [14]
2 years ago
6

18. In AABC, 44 is a right angle, and m4B 45°. What is the length of BC? If the answer is not an integer, leave it in simplest r

adical form. The
diagram is not drawn to scale.

054 ft

17√3 ft

17√2 ft

017

Mathematics
1 answer:
Lera25 [3.4K]2 years ago
8 0

Answer:

17√3

Step-by-step explanation:

In a 45-45-90 triangle, the hypotenuse is √3 times larger than the legs.

17*√3 = 17√3

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3 years ago
Round 12.5478 to the nearest hundredth's
sp2606 [1]

Answer:

I'm pretty sure it's 12.55

Step-by-step explanation:

4 0
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How long will it take an airplane flying at an average cruising speed of 230 miles per hour to fly 1380 miles?
lora16 [44]

Since it takes one hour to fly 230 miles, and we are looking for the number of hours, we can divide the number of miles we are given by the speed. In this case, we would divide 1380 by 230 to get 6. This means it would take 6 hours for the plane to fly 1380 miles.

7 0
3 years ago
Please help on the area for 7 8 9 please all i need help is
weeeeeb [17]

The area of the equilateral, isosceles and right angled triangle are 12.6mm², 9.61in² and 16.81yds² respectively.

<h3>What is the area of the equilateral, isosceles and right angle triangle?</h3>

Note that:

The area of an Equilateral triangle is expressed as A = ((√3)/4)a²

Where a is the dimension of the side.

The area of an Isosceles triangle is expressed as A = (ah)/2

Where a is the dimension of the base and h is the height.

The area of a Right angled triangle is expressed as A = (ab)/2

Where a and b is the dimension of the two sides other than the hypotenuse.

For the Equilateral triangle.

Given that;

  • a = 5.4mm
  • Area A = ?

A = ((√3)/4)(5.4mm)²

A = ((√3)/4)( 29.16mm² )

A = 12.6mm²

Area of the Equilateral triangle is 12.6mm²

For the Isosceles triangle.

Given that;

  • Base a = 3.4in
  • Slant height b = 5.9in
  • height h = ?
  • Area A = ?

The height h is the imaginary line drawn upward from the center of a.

First, we calculate the height using Pythagorean theorem

x² = y² + z²

Where x = b = 5.9in, y = a/2 = 3.4in/2 = 1.7in, and z = h

(5.9in)² = (1.7in)² + h²

34.81in² = 2.89in² + h²

h² = 34.81in² - 2.89in²

h² = 31.92in²

h = √31.92in²

h = 5.65in

Now, the area will be;

A = (ah)/2

A = (3.4in × 5.65in )/2

A = 19.21in²/2

A = 9.61in²

Area of the Isosceles triangle is 9.61in².

For the Right angled triangle

Given that;

  • a = 8.2yds
  • b = 4.1yds
  • c = 9.17yds
  • Area A = ?

A = (ab)/2

A = ( 8.2yds × 4.1yds)/2

A = ( 33.62yds²)/2

A = 16.81yds²

Area of the Right angled triangle is 16.81yds²

Therefore, the area of the equilateral, isosceles and right angled triangle are 12.6mm², 9.61in² and 16.81yds² respectively.

Learn more about Pythagorean theorem here: brainly.com/question/343682

#SPJ1

6 0
2 years ago
What is the average of 11, 16, 15, 22
Mazyrski [523]

<em><u>The average would be 16.</u></em>

To calculate this, add all the values together then divide that sum by the amount of data values.

11+16+15+22=64

64÷4=16

6 0
3 years ago
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