(-2+1^2)+5(12/3)-9
(1)+5(4)-9
1+20-9
21-9
12
SOLUTION
Given B is the midpoint of AC ABBC 1 and C is the midpoint of BD BCCD2 ie on adding 1 and 2 ART We get ABBCBCCD hence ABCD hence proved
Answer:
49/8 is the value of k
Step-by-step explanation:
We have the system
x = -2y^2 - 3y + 5
x=k
We want to find k such that the system intersects once.
If we substitute the second into the first giving us k=-2y^2-3y+5 we should see we have a quadratic equation in terms of variable y.
This equation has one solution when it's discriminant is 0.
Let's first rewrite the equation in standard form.
Subtracting k on both sides gives
0=-2y^2-3y+5-k
The discriminant can be found by evaluating
b^2-4ac.
Upon comparing 0=-2y^2-3y+5-k to 0=ax^2+bx+c, we see that
a=-2, b=-3, and c=5-k.
So we want to solve the following equation for k:
(-3)^2-4(-2)(5-k)=0
9+8(5-k)=0
Distribute:
9+40-8k=0
49-8k=0
Add 8k on both sides:
49=8k
Divide both sides by 8"
49/8=k
The lemght could be 15 because a triangle has two long sides and one short one and the two are equivalent
Answer:
x=2
Step-by-step explanation:
This is an isosceles triangle you can tell by the two congruent/equal angles
angle A is congruent/equal to angle C by definition the two sides will be equal.
x-4=3x-8
1. get all you x's on one side. by subtracting x from both sides you don't have to deal with dividing by negatives later on.
x-x-4=3x-x-8
-4=2x-8
now you want to get x by itself so add 8 to both sides
4= 2x
x is still not by itself so divide both sides by 2
x=2
