We know that the points at which the parabola intersects the x axis are
(-5,0) and (1,0)
So the extent between these two points would be the base of the triangle
lets find the length of the base using the distance formula
![\sqrt{[(-5-1)^{2}+(0-0)^{2} ]}](https://tex.z-dn.net/?f=%20%5Csqrt%7B%5B%28-5-1%29%5E%7B2%7D%2B%280-0%29%5E%7B2%7D%20%5D%7D%20%20)
the base b=6
We will get the height of the triangle when we put x=0 in the equation
y=a(0+5)(0-1)
y=-5a
so height = -5a (we take +5a since it is the height)
We know that the area of the triangle =
× 6 × (5a) = 12
15a=12
a= 
3,500 (.024) = 84
84 (5) = $420 in simple interest
Answer:
= 4sqrt3
Step-by-step explanation:
-2i sqrt -12
= -2i * i sqrt12 <-- i^2 = -1
= -2i^2sqrt12
= -2(-1)sqrt12
= 2sqrt12
= 2 * 2 sqrt3
= 4sqrt3
First, find out x
x+15+2x+15=180
3x+30=180
3x=150
x=50
so the two angels are x+15=65(let's name is ∠5 for convenience), and ∠6= 2x+15=115
notice the two inner lines are marked as congruent, so
∠4=∠5=65
∠1=180-∠4-∠5=180-65-65=50
Name the right bottom angle ∠7, ∠3=∠7 and ∠3+∠7=the exterior angle 100 degree, therefore, ∠3=50
∠2+∠3=∠4, therefore, ∠2=∠4-∠3=65-50=15
∠1=50, ∠2=15, ∠3=50, ∠4=65
