Point R is the midpoint of FE¯¯¯¯¯ , so the coordinates of point R are (3a, b).
In △DEF , the length of the base, DF¯¯¯¯¯ , is
4a, and the height is 2b, so its area is
1/2×4a×2b = 4ab.
In △QRP , the length of the base, QR¯¯¯¯¯ , is
3a-a = 2a, and the height is b, so its area is 1/2×2a×b = ab .
Comparing the expressions for the areas proves that the area of the triangle created by joining the midpoints of an isosceles triangle is one-fourth the area of the larger isosceles triangle.
258/65 = 3.9692
= 4.0 hours rounded to the nearest tenth hour
Step-by-step explanation:
The basic form of equation:
(x-h)²=4a(y-k),
(h,k)=coordinates of vertex
(h, k+a) = coordinate of focus
For given parabola:
axis of symmetry: x=2
(h, k) =(2,-3)
(h, k+a)=(2,5)
k+a=5
-3+a=5
a=8(distance from vertex to focus on the axis of symmetry)
equation: (x-2)²=4×8(y+3)
(x-2)²= 32(y+3)
Answer:
Yes. Her solution is correct.
Step-by-step explanation:
Let's check if Jenna solution is correct:
To solve the equation 2x^2 +5x - 42 = 0, we can use Bhaskara's formula:
D = b^2 - 4ac = 25 + 4*2*42 = 25+336 = 361
sqrt(D) = 19
x1 = (-5 + 19)/4 = 14/4 = 7/2
x2 = (-5 - 19)/4 = -24/4 = -6
We must agree with Jenna's solution, because the values she found as solution are correct: with we replace these values of x in the equation, we will find 0 = 0, which is correct and proves that these values are the solution of the equation.
