Answer:
Pythagorean Theorem: c2 = a2 + b2
Find the area by adding the areas of the three triangles. The area of a right triangle is: A = ½bh
Two triangles are identical so you can just multiply the area of the first triangle by two: 2A1 = 2(½bh) = 2(½ab) = ab.
The total area of the trapezoid is : A1 + A2 = ab + ½c2
You multiply both sides by 2 to get rid of the ½: (a2 + 2ab + b2) = 2ab + c2
You subtract out the 2ab: a2 + b2 = c2.
Then what is left is the proof: a2 + b2 = c2
<span><u><em>The correct answers are: </em></u>
1) A;
2) A;
4) D
3 cannot be done because the graph is not shown.
<u><em>Explanation:</em></u><span><u><em> </em></u>
1) Shifting a graph to the left, we would normally think of subtracting 1 from the function. However, horizontal translations are the opposite; left means adding 1 to x, while right means subtracting 1 from x.
2) A reflection in the x-axis means the y-coordinate will be negated (the opposite sign). This means that g(x)=-f(x)=-(x^2+5=-x^2-5.
4) To perform a vertical stretch of a function, we multiply by the factor; this gives us y=6x.</span></span>
=Y2-10Y
We move all terms to the left:
-(Y2-10Y)=0
We add all the numbers together, and all the variables
-(+Y^2-10Y)=0
We get rid of parentheses
-Y^2+10Y=0
We add all the numbers together, and all the variables
-1Y^2+10Y=0
a = -1; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-1)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
Y1=−b−Δ√2aY2=−b+Δ√2a
Δ‾‾√=100‾‾‾‾√=10
Y1=−b−Δ√2a=−(10)−102∗−1=−20−2=+10
Y2=−b+Δ√2a=−(10)+102∗−1=0−2=0
