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rjkz [21]
4 years ago
15

Please see the picture...

Mathematics
1 answer:
kow [346]4 years ago
5 0
What's there to answer homes
You might be interested in
I could probably figure this out but I have way too much homework so It'd be nice to finish this pathway.
pentagon [3]

Yeah that's right always be a hardworking person

8 0
3 years ago
A plane is at cruising altitude of 10,200 feet. If the angle of depression to the run way is 18 degrees, what is the distance, t
MatroZZZ [7]

Answer: 33,007.9\ ft

Step-by-step explanation:

Given

The plane is cruising at an altitude of h=10,200\ ft

The angle of depression is \theta=18^{\circ}

Suppose the distance of the plane to the runway is x

from figure

\Rightarrow \sin 18^{\circ}=\dfrac{h}{x}\\\\\Rightarrow \sin 18^{\circ}=\dfrac{10,200}{x}\\\\\Rightarrow x=33,007.9\ ft

4 0
3 years ago
What should be added to x^2 + 4 to get 2x^2 + 5
Art [367]

Answer:

x+1

Step-by-step explanation:

4 0
3 years ago
Am i correct? calculus
Advocard [28]
Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer.  That matters because the derivative of a constant is zero.

now, those are the values when the rocket is 30 feet up above.

\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
\\\\\\
\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
\\\\\\
\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
-------------------------------\\\\


\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
-------------------------------\\\\
\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
\\\\\\
\cfrac{d\theta }{dt}=\cfrac{11}{75}

8 0
3 years ago
I'LL GIVE YOU A LOT OF POINTS !!!! Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use
Mariana [72]

Answer:

For the perimeters, x must be equal to 2.

For the areas, it is either undefined, or something.

Step By Step Explanation:

You can first find the perimeters for both sides.

For the left shape, we add the two sides of 6 and x + 4 to get x + 10.

Then we multiply x + 10 by 2 because there are 4 sides, and we only got 2 sides.

The perimeter of the first shape is 2x + 20.

The second shape can be solved by doing the same thing by adding 2 and 3x + 4 to get 3x + 6.

3x + 6 times 2 is 6x + 12.

The second perimeter is 6x + 12.

If both sides are supposed to be equal, then we can write these two expressions we solved for like:

6x + 12 = 2x + 20.

Subtraction property of equality

6x + 12 - 12 = 2x + 20 - 12

Simplify

6x = 2x + 8

Again

6x - 2x = 2x - 2x + 8

Simplify

4x = 8

Division property of equality

4/4x = 8/4

Simplify

x = 2

So if x = 2, the perimeters will be the same.

You can confirm this by plugging it back into either equation.

For the areas, we just multiply the length and width for both shapes, so we get

6(x+4)  =  2(3x+4)

Since they are supposed to be equal.

We simplify and get

6x + 24 = 6x + 8

We know this is false and is not possible, since we can remove the 6x because it is on both sides.

We also know that 24 is not equal to 8 (who thought!)

:D

24 ≠ 8

So it is undefined or whatever you call it.

6 0
3 years ago
Read 2 more answers
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