What set of transformations are applied to parallelogram ABCD to create A'B'C'D'?
1 answer:
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The college student took 13 courses of 3 credit hours.
Step-by-step explanation:
Given,
Total credits = 59
Total courses = 18
Let,
Number of 3 credit hour courses = x
Number of 4 credit hour courses = y
According to given statement;
x+y=18 Eqn 1
3x+4y=59 Eqn 2
We will eliminate y to find the number of 3 credit hour courses, therefore,
Multiplying Eqn 1 by 4
Subtracting Eqn 2 from Eqn 3
The college student took 13 courses of 3 credit hours.
Keywords: linear equation, subtraction
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One thing to bear in mind with trigonometric expressions is that, the exponent location is next to the function name, instead of the whole expression, though it applies to the whole thing, namely cos²(x), is really [ cos(x) ]², and that matters for using the chain rule.
![\bf y=cos^2(x)-sin^2(x)\implies y=[cos(x)]^2-[sin(x)]^2 \\\\\\ \cfrac{dy}{dx}=\stackrel{chain~rule}{2cos(x)\cdot -sin(x)}~~-~~\stackrel{chain~rule}{2sin(x)\cdot cos(x)} \\\\\\ \cfrac{dy}{dx}=-2cos(x)sin(x)-2cos(x)sin(x) \\\\\\ \cfrac{dy}{dx}=-4cos(x)sin(x)](https://tex.z-dn.net/?f=%5Cbf%20y%3Dcos%5E2%28x%29-sin%5E2%28x%29%5Cimplies%20y%3D%5Bcos%28x%29%5D%5E2-%5Bsin%28x%29%5D%5E2%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D%5Cstackrel%7Bchain~rule%7D%7B2cos%28x%29%5Ccdot%20-sin%28x%29%7D~~-~~%5Cstackrel%7Bchain~rule%7D%7B2sin%28x%29%5Ccdot%20cos%28x%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D-2cos%28x%29sin%28x%29-2cos%28x%29sin%28x%29%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D-4cos%28x%29sin%28x%29)