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3 years ago

A jet flies at an altitude of 52,800 feet. What is the jets height in miles?

Mathematics

1 answer:

WITCHER [35]3 years ago

5 0

Hi, I have an answer for you, hope is O.K.

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Find the value of the power. Please do all four questions for 15 pts #1 6³ #2 9² #3 3⁴ #4 18²

defon

Answer:

6³ = 216

9² = 81

3⁴ = 81

18² = 324

Step-by-step explanation:

6³ = 6 · 6 · 6 = 36 · 6 = 216

9² = 9 · 9 = 81

3⁴ = 3 · 3 · 3 · 3 = 9 · 3 · 3 = 27 · 3 = 81

18² = 18 · 18 = 324

3 0

3 years ago

I NEED HELP PLEASE !!!! ;(

marissa [1.9K]

Answer:

The answer is always

Hope this helps......... :)

7 0

3 years ago

Farmer Jones owns a triangular plece of land.

Elenna [48]

155 is your answer lmk if you need anything else

6 0

3 years ago

Read 2 more answers

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

Read 2 more answers

Write and simplify an expression for the volume of a rectangular prism with length ℓ m, width 5 m, and height 6.5 m. What is the

Dominik [7]

Step-by-step explanation:

V= 32.5ℓ (m^3)

when ℓ= 8.4 m , V= 273 m^2

4 0

2 years ago