Answer:
Theorem: The diagonals of a parallelogram bisect each other. Proof: Given ABCD, let the diagonals AC and BD intersect at E, we must prove that AE ∼ = CE and BE ∼ = DE. The converse is also true: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.
Step-by-step explanation:
Answer:
just multiply 35 by x and then 10 by however many lawns he mowed (if that makes sense)
Step-by-step explanation:
Answer:
c
Step-by-step explanation:
I think you missed attaching the photo, please see my attachment.
And the correct answer is C,
When you look at where the arc meets the parallel lines, if you create a seam between two points, you get a straight line parallel to the horizontal lines so it makes the corresponding angles are congruent.
1. Let a and b be coefficients such that

Combining the fractions on the right gives



so that

2. a. The given ODE is separable as

Using the result of part (1), integrating both sides gives

Given that y = 1 when x = 1, we find

so the particular solution to the ODE is

We can solve this explicitly for y :


![\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|](https://tex.z-dn.net/?f=%5Cln%7Cy%7C%20%3D%20%5Cln%5Cleft%7C%5Csqrt%5B3%5D%7B%5Cdfrac%7B5x%7D%7B2x%2B3%7D%7D%5Cright%7C)
![\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}](https://tex.z-dn.net/?f=%5Cboxed%7By%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B5x%7D%7B2x%2B3%7D%7D%7D)
2. b. When x = 9, we get
![y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B45%7D%7B21%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B15%7D7%7D%20%5Capprox%20%5Cboxed%7B1.29%7D)