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prohojiy [21]
3 years ago
10

Given: BE bisects AD at C, AB ⊥ BC , DE ⊥ EC, AB ≅ DE Prove: ABC ≅DEC

Mathematics
1 answer:
faltersainse [42]3 years ago
6 0

Answer:

$

Step-by-step explanation:

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Find f(x) and g(x) so that the function can be described as y = f(g(x)). (5 points) y = eight divided by square root of quantity
zubka84 [21]

Answer:

  • f(x) = 8/√x
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Step-by-step explanation:

For this sort of problem, there can be an infinite number of answers.

It can be convenient to choose one of the simpler answers by looking at the operations that are performed on the variable. Here, you have ...

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You can work from the bottom up and define the outer function (f(x)) to be any of these operations. In our answer above, we have elected to include the "square root" and the "8 divided by that root" in our definition of f.

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6 0
3 years ago
Question 18 of 25<br> What is tan 45°?
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Answer:

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5 0
2 years ago
Find the length of the tangent segment AB to the circles centered at O and O' whose radii are a and b respectively when the circ
bixtya [17]

Answer:

AB=2\sqrt{ab}

Step-by-step explanation:

From figure,

OA=a,  \quad \quad O'B=b\\\Rightarrow OO'= (a+b) \quad \quad \text{and}\quad OD=(a-b)

In triangle OO'D

  (OO')^2=(OD)^2+(O' D)^2

\Rightarrow (a+b)^2=(a-b)^2+(O' D)^2\\\Rightarrow a^2+b^2+2ab-a^2-b^2+2ab=(O' D)^2\\\Rightarrow 4ab=(O' D)^2\\\Rightarrow O'D=2\sqrt{ab} \\\Rightarrow O' D=2\sqrt{ab}=AB \quad \quad [\because O' DAB\;\; \text{is a rectangle.}]

Hence, AB=2\sqrt{ab}

6 0
3 years ago
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