A hiker is standing on the ground looking up to the top of a tree with an angle of elevation of °= 55 The situation represented
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
1. Swap sides
Swap sides:
2. Isolate the y
Multiply to both sides by 18:
Group like terms:
Simplify the fraction:
Multiply the fractions:
Simplify the arithmetic:
---------------------------
Why learn this:
- Linear equations cannot tell you the future, but they can give you a good idea of what to expect so you can plan ahead. How long will it take you to fill your swimming pool? How much money will you earn during summer break? What are the quantities you need for your favorite recipe to make enough for all your friends?
- Linear equations explain some of the relationships between what we know and what we want to know and can help us solve a wide range of problems we might encounter in our everyday lives.
---------------------
Terms and topics
- Linear equations with one unknown
The main application of linear equations is solving problems in which an unknown variable, usually (but not always) x, is dependent on a known constant.
We solve linear equations by isolating the unknown variable on one side of the equation and simplifying the rest of the equation. When simplifying, anything that is done to one side of the equation must also be done to the other.
An equation of:
in which and
are the constants and
is the unknown variable, is a typical linear equation with one unknown. To solve for
in this example, we would first isolate it by subtracting
from both sides of the equation. We would then divide both sides of the equation by
resulting in an answer of:
It looks like the integral is
where <em>C</em> is the circle of radius 2 centered at the origin.
You can compute the line integral directly by parameterizing <em>C</em>. Let <em>x</em> = 2 cos(<em>t</em> ) and <em>y</em> = 2 sin(<em>t</em> ), with 0 ≤ <em>t</em> ≤ 2<em>π</em>. Then
Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on <em>C</em> nor in the region bounded by <em>C</em>, so
where <em>D</em> is the interior of <em>C</em>, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result: .