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sweet [91]
3 years ago
11

Given: ∠AOB is a central angle and ∠ACB is a circumscribed angle. Prove: △ACO ≅ △BCO Circle O is shown. Line segments A O and B

O are radii. Tangents C B and C B intersect at point C outside of the circle. A line is drawn to connect points C and O. We are given that angle AOB is a central angle of circle O and that angle ACB is a circumscribed angle of circle O. We see that AO ≅ BO because . We also know that AC ≅ BC since . Using the reflexive property, we see that . Therefore, we conclude that △ACO is congruent to △BCO by the .
Mathematics
2 answers:
charle [14.2K]3 years ago
6 0

Answer:

all radii of the same circle are congruent

tangents to a circle that intersect are congruent

side CO is congruent to side CO

SSS congruency theorem

Step-by-step explanation:

m_a_m_a [10]3 years ago
6 0

Answer:

1,3,3,1

Step-by-step explanation:

Trust the process

edge 2021

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i have a problem im struggling with it says. is 6 a solution to the equation x2 - 4=5x explain why or why not?
Eddi Din [679]

Answer:

No, 6 is not a solution to the equation x2-4=5x

Step-by-step explanation:

6(2)-4=5(6)

12-4=30

8=30

The question is basically asking if you replace x with 6, will the equation equal each other on both sides, for example 6=6 or 3.5=3 1/2. In this case the equation gives us 8=30 which means 6 would not be a solution since both sides aren't equal to one another, 30 is bigger than 8.

6 0
4 years ago
Match the base to the corresponding height. Base (b) Height (h) h h b b h
Musya8 [376]

Answer:2 b

Step-by-step explanation:3 h

8 0
3 years ago
Read 2 more answers
Given f(x), find g(x) and h(x) such that f(x)=g(h(x)) and neither g(x) nor h(x) is solely x. f(x)=−3x+4⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3+5
iVinArrow [24]

Answer:

h(x)=-2x+4

g(x)=\sqrt[3]{x}-5

Step-by-step explanation:

Consider the given function is

f(x)=\sqrt[3]{-2x+4}-5

It is given that f(x)=g(h(x)) and neither g(x) nor h(x) is solely x.

f(x)=\sqrt[3]{(-2x+4)}-5

Let h(x)=-2x+4, then we get

f(x)=g(h(x))=\sqrt[3]{h(x)}-5

Substitute h(x)=x in the above function.

g(x)=\sqrt[3]{x}-5

Therefore, the required functions are h(x)=-2x+4 and g(x)=\sqrt[3]{x}-5.

Check the solutions.

g(h(x))=g(-2x+4)             [\because h(x)=-2x+4]

g(h(x))=\sqrt[3]{-2x+4}-5            [\because g(x)=\sqrt[3]{x}-5]

g(h(x))=f(x)

Therefore, our solution is correct.

7 0
4 years ago
Which is the completely factored form of 12x4 + 39x3 + 9x2?
svetoff [14.1K]

Answer:

The second option, 3x^2(x+3)(4x+1)

Step-by-step explanation:

Factor out the common term, which is 3x^2

This then becomes 3x^2(4x^2+13x+3)

Factor the inside term and it becomes

(3x^2)(x+3)(4x+1)

3 0
2 years ago
Solve 2x+y=7 for y what is y ?
masha68 [24]
You cant solve y but you can form an equation in terms of y
which is y=7-2x

if you were given another equation that has the term y and x than you can solve it.also that is a simultaneous equation.

Hope this helps,dont forget to put this as the brainliest answer if it does to help me out too. xD
6 0
3 years ago
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