Answer:
3141.59
Step-by-step explanation:
Answer: 
<u>Step-by-step explanation:</u>
Isolate w by performing the following steps
- Multiply by 6 on both sides to clear the denominator
- Subtract 3 from both sides
- Divide both sides by 2
![y=\dfrac{1}{2}+\dfrac{w}{3}\\\\\\6\bigg[y=\dfrac{1}{2}+\dfrac{w}{3}\bigg]\quad \implies \quad 6y=3+2w\\\\\\6y-3=3-3+2w\quad \implies \quad 6y-3=2w\\\\\\\dfrac{6y-3}{2}=\dfrac{2w}{2}\quad \implies \quad \large\boxed{\dfrac{6y-3}{2}=w}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7Bw%7D%7B3%7D%5C%5C%5C%5C%5C%5C6%5Cbigg%5By%3D%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7Bw%7D%7B3%7D%5Cbigg%5D%5Cquad%20%5Cimplies%20%5Cquad%206y%3D3%2B2w%5C%5C%5C%5C%5C%5C6y-3%3D3-3%2B2w%5Cquad%20%5Cimplies%20%5Cquad%206y-3%3D2w%5C%5C%5C%5C%5C%5C%5Cdfrac%7B6y-3%7D%7B2%7D%3D%5Cdfrac%7B2w%7D%7B2%7D%5Cquad%20%5Cimplies%20%5Cquad%20%5Clarge%5Cboxed%7B%5Cdfrac%7B6y-3%7D%7B2%7D%3Dw%7D)
Answer:
For Lin's answer
Step-by-step explanation:
When you have a triangle, you can flip it along a side and join that side with the original triangle, so in this case the triangle has been flipped along the longest side and that longest side is now common in both triangles. Now since these are the same triangle the area remains the same.
Now the two triangles form a quadrilateral, which we can prove is a parallelogram by finding out that the opposite sides of the parallelogram are equal since the two triangles are the same(congruent), and they are also parallel as the alternate interior angles of quadrilateral are the same. So the quadrilaral is a paralllelogram, therefore the area of a parallelogram is bh which id 7 * 4 = 7*2=28 sq units.
Since we already established that the triangles in the parallelogram are the same, therefore their areas are also the same, and that the area of the parallelogram is 28 sq units, we can say that A(Q)+A(Q)=28 sq units, therefore 2A(Q)=28 sq units, therefore A(Q)=14 sq units, where A(Q), is the area of triangle Q.
Answer:
This is just a relation.
Step-by-step explanation:
It is easy to see if a set of ordered pairs is a relation or a relation that is a function.
If all the points are distinct, then you just have to look at the 
-coordinates.
If all the -coordinates are different, then it is a function. Otherwise, it is a just a relation.
So this cannot be a function because there is at least two different points that have 
.
It will take 6.875h for the second car to overtake the first...
