7x1=7, then 4x1/4= 4x1=4 divided by 4= 1. Then subtract 7- 1= 6
Remember (a²-b²)=(a-b)(a+b)
solve for a single variable
solve for y in 2nd
add y to both sides
x²-7=y
sub (x²-7) for y in other equaiton
4x²+(x²-7)²-4(x²-7)-32=0
expand
4x²+x⁴-14x²+49-4x²+28-32=0
x⁴-14x²+45=0
factor
(x²-9)(x²-5)=0
(x-3)(x+3)(x-√5)(x+√5)=0
set each to zero
x-3=0
x=3
x+3=0
x=-3
x-√5=0
x=√5
x+√5=0
x=-√5
sub back to find y
(x²-7)=y
for x=3
9-7=2
(3,2)
for x=-3
9-7=2
(-3,2)
for √5
5-7=-2
(√5,-2)
for -√5
5-7=-2
(-√5,-2)
the intersection points are
(3,2)
(-3,2)
(√5,-2)
(-√5,-2)
H is the altitude of a right triangle, so h²=AD*DC=(20-4)*4=64
h=√64=8
D is the correct answer.
Answer:
A) 
B)
Step-by-step explanation:
AB has length a and is divided by points P and Q into AP , PQ , and QB , such that AP = 2PQ = 2QB
A) Therefore, AP = 2QB
QB = AP/2
The midpoint of QB = QB/2 = (AP/2)/2 = AP/4
AP = 2PQ, Therefore PQ = AP/2
Since the length of AB = a
AB = AP + PQ + QB = a
AP + AP/2 + AP/2 = a
AP + AP = a
2AP = a
AP = a/2
The distance between point A and the midpoint of segment QB = AP + PQ + QB/2 = AP + AP/2 + AP/4 = 7/4(AP)
But AP = a/2
Therefore The distance between point A and the midpoint of segment QB = 7/4(a/2)= 
B)
the distance between the midpoints of segments AP and QB = AP/2 + PQ + QB/2 = AP/2 + AP/2 + AP/4 = 5/4(AP)
But AP = a/2
Therefore the distance between the midpoints of segments AP and QB = 5/4(AP) = 
Answer:

Step-by-step explanation:
1) Flip the second fraction and change the operation to multiplication:

2) Cross cancel factors:

3) Multiply:

4) Simplify:
