9514 1404 393
Answer:
x = 57
Step-by-step explanation:
Where parallel lines are crossed by a transversal, all obtuse angles are congruent and supplementary to all acute angles, which are also congruent.
x = 180 -123
x = 57
11^0 is 1.
11^2= 121
121/1= 121
Final answer: 121
<h3>
Answer: Vertical angles</h3>
Explanation:
Ignore the ray that angles 2 and 3 are adjacent to. It might help to copy the drawing but erase that ray in question. All we really care about is the X shape the two lines form along with angles 1 and 4. These two angles are opposite one another. We consider them vertical angles. They don't have to align vertically. They simply have to be opposite one another like this in this X configuration. Vertical angles are always congruent to one another.
Answer:A??
Step-by-step explanation:
The values of h and k when f(x) = x^2 + 12x + 6 is in vertex form is -6 and -30
<h3>How to rewrite in vertex form?</h3>
The equation is given as:
f(x) = x^2 + 12x + 6
Rewrite as:
x^2 + 12x + 6 = 0
Subtract 6 from both sides
x^2 + 12x = -6
Take the coefficient of x
k = 12
Divide by 2
k/2 = 6
Square both sides
(k/2)^2 = 36
Add 36 to both sides of x^2 + 12x = -6
x^2 + 12x + 36= -6 + 36
Evaluate the sum
x^2 + 12x + 36= 30
Express as perfect square
(x + 6)^2 = 30
Subtract 30 from both sides
(x + 6)^2 -30 = 0
So, the equation f(x) = x^2 + 12x + 6 becomes
f(x) = (x + 6)^2 -30
A quadratic equation in vertex form is represented as:
f(x) = a(x - h)^2 + k
Where:
Vertex = (h,k)
By comparison, we have:
(h,k) = (-6,-30)
Hence, the values of h and k when f(x) = x^2 + 12x + 6 is in vertex form is -6 and -30
Read more about quadratic functions at:
brainly.com/question/1214333
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