Answer:
The answer to your question is: first piece = 4 in, second piece = 8 in, third piece = 12 in
Step-by-step explanation:
Data
total length = 24 inches
first piece = x in
second piece = 2x
third piece = 3x
Equation
x + 2x + 3x = 24
6x = 24
x = 24/6
x = 4 length of the first piece
second piece = 2(4) = 8 inches
third piece = 3(4) = 12 inches
Formula:
A(t)=P(1+(r/n))^)nt)
A(4)=2500(1+(0.04/4))^(4*4)
A(4) = 2500(1.01)^16
A(4) = 2500*1.1726
<span>A(4) = $2931.45</span>
Answer: 21.375 cm2
Step-by-step explanation:
Area of rectangle = length x width = 3.5 x 4.4 = 15.75 cm2
Area of triangle = 1/2 x base x height = 1/2 x 2.5 x 4.5 = 5.625 cm2
Add rectangle and triangle area = 21.375 cm2
Answer:

Step-by-step explanation:
Slant Height = 10 cm
Base Edge = 12 cm
<u>Finding the Height using Pythagorean Theorem:</u>
![\sf c^2 = a^2 + b^2 \\\\Where \ c = 10\ cm , b = 6\ cm [Half of 12] \ and \ a \ is \ height\\\\10^2 = a^2 + 6^2\\\\100=a^2 +36\\\\100-36 = a^2\\\\64 = a^2\\\\Taking \ sqrt \ on \ both \ sides\\\\Height= 8 \ cm](https://tex.z-dn.net/?f=%5Csf%20c%5E2%20%3D%20a%5E2%20%2B%20b%5E2%20%5C%5C%5C%5CWhere%20%5C%20c%20%3D%2010%5C%20cm%20%2C%20b%20%3D%206%5C%20cm%20%5BHalf%20of%2012%5D%20%5C%20and%20%5C%20a%20%5C%20is%20%5C%20height%5C%5C%5C%5C10%5E2%20%3D%20a%5E2%20%2B%206%5E2%5C%5C%5C%5C100%3Da%5E2%20%2B36%5C%5C%5C%5C100-36%20%3D%20a%5E2%5C%5C%5C%5C64%20%3D%20a%5E2%5C%5C%5C%5CTaking%20%5C%20sqrt%20%5C%20on%20%5C%20both%20%5C%20sides%5C%5C%5C%5CHeight%3D%208%20%5C%20cm)
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
<u>Now, Finding the volume:</u>

![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h2>~AnonymousHelper1807</h2>
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.