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Pavlova-9 [17]
3 years ago
14

A+B=44 A : B is equvililent to 13 :8 Please solve

Mathematics
2 answers:
lilavasa [31]3 years ago
6 0
13:18 is in simplest form, so it’s multiplied by something, let’s call it x

So A + B equals 13x + 8x

13x + 8x = 44
21x = 44
x =44/21

Plug in to check:
13(44/21) + 8(44/21) = 44
572/21 + 352/21 = 44
924/21 = 44
44 = 44 :)
Veronika [31]3 years ago
3 0

Step-by-step explanation:

<h3>13:8</h3><h3>A=13x. B=8x</h3><h3>13x + 8x = 44</h3><h3> 21x=44</h3><h3> x=2.095</h3>
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A scale model of a tower is 24 inches tall and 18 inches wide. If the height of the actual tower is 60 feet, what is its width?
alex41 [277]

Answer:

45 feet

Step-by-step explanation:

To find the width of the model, we need to find the scale of the model to the actual tower.

Since we know the height of both towers, we can use that as the basis.

24 inches : 60 feet

12 inches : 30 feet

6 inches : 15 feet

So the width of the tower is 18 inches wide.

18 inches will then be equal to 12 + 7 inches : 30 + 15 feet

18 inches : 45 feet

6 0
3 years ago
If f(x) = x^2 + 8x – 2 and f(0), then the result is:<br><br> -2.<br> 0.<br> 8.
Anastasy [175]

Answer:

-2.

Step-by-step explanation:

f(0) = 0^2 + 8(0) - 2

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2 years ago
15 acres decreased by 80%
lawyer [7]
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4 0
3 years ago
A helicopter is flying above a town. the local high school is directly to the east of the helicopter at a 20° angle of depressio
guapka [62]
Answer: 4.5 miles

Explanation:

When you draw the situation you find two triangles.

1) Triangle to the east of the helicopter

a) elevation angle from the high school to the helicopter = depression angle from the helicopter to the high school = 20°

b) hypotensue = distance between the high school and the helicopter

c) opposite-leg to angle 20° = heigth of the helicopter

d) adyacent leg to the angle 20° = horizontal distance between the high school and the helicopter = x

2) triangle to the west of the helicopter

a) elevation angle from elementary school to the helicopter = depression angle from helicopter to the elementary school = 62°

b)  distance between the helicopter and the elementary school = hypotenuse

c) opposite-leg to angle 62° = height of the helicopter

d) adyacent-leg to angle 62° = horizontal distance between the elementary school and the helicopter = 5 - x

3) tangent ratios

a) triangle with the helicpoter and the high school

tan 20° = Height / x ⇒ height = x tan 20°

b) triangle with the helicopter and the elementary school

tan 62° = Height / (5 - x) ⇒ height = (5 - x) tan 62°

c) equal the height from both triangles:

x tan 20° = (5 - x) tan 62°

x tan 20° = 5 tan 62° - x tan 62°

x tan 20° + x tan 62° = 5 tan 62°

x  (tan 20° + tan 62°) = 5 tan 62°

⇒ x = 5 tant 62° / ( tan 20° + tan 62°)

⇒ x = 4,19 miles

=> height = x tan 20° = 4,19 tan 20° = 1,525 miles

4) Calculate the hypotenuse of this triangle:

hipotenuese ² = x² + height ² = (4.19)² + (1.525)² = 19.88 miles²

hipotenuse = 4.46 miles

Rounded to the nearest tenth = 4.5 miles

That is the distance between the helicopter and the high school.
5 0
3 years ago
Read 2 more answers
All the given quadrilaterals in the picture on the right are squares and #1 ≅ #3, #2 ≅ # 4. find the area of the shaded region,
Anika [276]

Answer:

Step-by-step explanation:

If you have a square of side l, its diagonal would be l\sqrt{2}, and its area l^2

If the big square has a area of 900, this implies that its side is \sqrt{900}, so the two diagonal of squares 2 and 4 added together would be \sqrt{900}, therefore one diagonal wold be \frac{\sqrt{900}}{2}. and its side \frac{\sqrt{900}}{2}\frac{1}{\sqrt{2}}. The area (of one square) is (\frac{\sqrt{900}}{2}\frac{1}{\sqrt{2}})^2=\frac{225}{2}

finally the two areas combined (squares 2 and 4) would be 225

6 0
2 years ago
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