AC is a tangent so by definition, it touches the circle at exactly one point (point C) and forms a right angle at the tangency point. So angle ACO is 90 degrees
The remaining angle OAC must be 45 degrees because we need to have all three angles add to 180
45+45+90 = 90+90 = 180
Alternatively you can solve algebraically like so
(angle OAC) + (angle OCA) + (angle COA) = 180
(angle OAC) + (90 degrees) + (45 degrees) = 180
(angle OAC) + 90+45 = 180
(angle OAC) + 135 = 180
(angle OAC) + 135 - 135 = 180 - 135
angle OAC = 45 degrees
Side Note: Triangle OCA is an isosceles right triangle. It is of the template 45-45-90.
Answer:The answer is x = -203.
Step-by-step explanation:
1. reduce brackets
2. reduce 10 - 523 to -513
3. reduce -513 x -2 to 1026
4. subtract 1026 from both sides
5. reduce 823 - 1026 to -203
Answer:
(5a−3)^2
Step-by-step explanation:
25a^2 - 30a + 9
Factor the expression by grouping. First, the expression needs to be rewritten as 25a^2+pa+qa+9. To find p and q, set up a system to be solved.
p+q=−30
pq=25×9=225
Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 225.
−1,−225
−3,−75
−5,−45
−9,−25
−15,−15
Calculate the sum for each pair.
−1−225=−226
−3−75=−78
−5−45=−50
−9−25=−34
−15−15=−30
The solution is the pair that gives sum −30.
p=−15
q=−15
Rewrite 25a^2 - 30a + 9 as (25a^2−15a)+(−15a+9).
(25a^2−15a)+(−15a+9)
Factor out 5a in the first and −3 in the second group.
5a(5a−3)−3(5a−3)
Factor out common term 5a−3 by using distributive property.
(5a−3)(5a−3)
Rewrite as a binomial square.
(5a−3)^2
Answer:
i Think its 1
Step-by-step explanation:
