Answer:
B
Step-by-step explanation:
So far, we know that:
∠D = ∠J.
And that:
DE:JK = 14:7 = 2:1
So, to prove that ΔDEF ~ ΔJKL by SAS, DF must be similar to JL, as those are the sides between the angle.
So:
DF:JL = 2:1.
Our answer is B.
F(x) = 2x² - 8x - 10.
This is a parabola open upward (since a>0) with an axis of symmetry = -b/2a:
a) axis of symmetry: x = -(-8)/(2*2) = 8/4 = 2. Then x = 2, which is the x component of the vertex
b) for x = 2, f(x) = f(2) = - 18 (component of y of the vertex)
c) VERTEX(2, - 18)
d) DISCRIMINENT: b² - 4.a.c = 64 - 4*2*(-10) = 144
BECAUSE THERE IS A Y AXIS AND AN X AXIS THOSE R THE 2 GIVIN LINES
*If you need me to do #5, DM me!
3. The area of a triangle can be given by (just plug and chug as always):

The area of the triangle is
6ft².
4. I will divide into a triangle and a rectangle (because the actual equation for the area of a pentagon requires it to be a perfect pentagon). Let's do the triangle first (height is 3 because you subtract 12 from 15):


Now we just add them:

So, the area of that pentagon is
108m².
5. You are actually wrong on this one because the area of a triangle is:

So, just halve your answer and it will be correct.
6. We can just split it into 4 triangles of equal area and then multiply the area of 1 triangle by 4 to get the total area. Let's do just that:

Multiply by 4 to get total area:

So, the area of the given rhombus is
25cm².