Answer:
the 2nd one and the 3rd one
Step-by-step explanation:
hope this helps
From your previous questions, you know
(3<em>w</em> + <em>w</em>⁴)' = 3 + 4<em>w</em>³
(2<em>w</em>² + 1)' = 4<em>w</em>
So by the quotient rule,
<em>R'(w)</em> = [ (2<em>w</em>² + 1)•(3<em>w</em> + <em>w</em>⁴)' - (3<em>w</em> + <em>w</em>⁴)•(2<em>w</em>² + 1)' ] / (2<em>w</em>² + 1)²
That is, the quotient rule gives
<em>R'(w)</em> = [ (denominator)•(derivative of numerator) - (numerator)•(derivative of denominator) ] / (denominator)²
I'm not entirely sure what is meant by "unsimplified". Technically, you could stop here. But since you already know the component derivatives, might as well put them to use:
<em>R'(w)</em> = [ (2<em>w</em>² + 1)•(3 + 4<em>w</em>³) - (3<em>w</em> + <em>w</em>⁴)•(4<em>w</em>) ] / (2<em>w</em>² + 1)²
Solve 89 = .7x and you will get 127.14
Y=-1
X= 3
This is the answer. If u want to see the work t is on the photo being attached
∠TAC = ∠ABC (The angle between the chord and the tangent is equal to the angle in the alternate segment)
Since, EDCB is a cyclic quadrilateral,
∠CBD + ∠CED = 180 ----- I
- ∠CED + ∠AED = 180 -----II
----------------------------------------------------
∠ CBD = ∠AED
Hence proved
To learn more about circle geomtery, refer to brainly.com/question/24375372
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