Answer:
a
Step-by-step explanation:
The answer to your question is b
You can solve this by using "similar triangles".
In triangle ABC, we are looking for side AC which is x. Side AC is similar to side DF in triangle EDF.
You can solve for side x by picking two sides in triangle ABC and their corresponding sides in triangle EDF. This is what I mean:

Substitute for the values of AC, BC, DF and EF:


To solve for y, do the same thing. Pick two sides on triangle ABC and their corresponding sides in triangle DEF.

Substitute for the values and solve:


We have the value x to be 5.5 units and y to be 6 units.
Answer:
2: 99° 3: 81° 4: 81°
Step-by-step explanation:
2 is supplementary to 1 so subtract 81 from 180 = 99.
3 is vertical angle with 1 so they are equal
4 is alternate interior angle with 3 so they are equal
hope this helps stay safe :)
Answer:
Step-by-step explanation:
Given a general quadratic formula given as ax²bx+c = 0
To generate the general formula to solve the quadratic equation, we can use the completing the square method as shown;
Step 1:
Bringing c to the other side
ax²+bx = -c
Dividing through by coefficient of x² which is 'a' will give:
x²+(b/a)x = -c/a
- Completing the square at the left hand side of the equation by adding the square of half the coefficient x i.e (b/2a)² and adding it to both sides of the equation we have:
x²+(b/a)x+(b/2a)² = -c/a+(b/2a)²
(x+b/2a)² = -c/a+(b/2a)²
(x+b/2a)² = -c/a + b²/4a²
- Taking the square root of both sides
√(x+b/2a)² = ±√-c/a + b²/√4a²
x+b/2a = ±√(-4ac+b²)/√4a²
x+b/2a =±√b²-4ac/2a
- Taking b/2a to the other side
x = -b/2a±√√b²-4ac/2a
Taking the LCM:
x = {-b±√b²-4ac}/2a
This gives the vertex form with how it is used to Solve a quadratic equation.