Answer:
X = 0 , y=2
Step-by-step explanation:
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Y = (x + 2)(x − 3) Find the zero of the function
1 answer:
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The nth taylor polynomial for the given function is
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Given:
f(x) = ln(x)
n = 4
c = 3
nth Taylor polynomial for the function, centered at c
The Taylor series for f(x) = ln x centered at 5 is:
Since, c = 5 so,
Now
f(5) = ln 5
f'(x) = 1/x ⇒ f'(5) = 1/5
f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25
f'''(x) = 2/x³ ⇒ f'''(5) = 2/5³ = 2/125
f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625
So Taylor polynomial for n = 4 is:
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Hence,
The nth taylor polynomial for the given function is
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
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*Correct Question:
Based on the similar triangles shown below, Theodore claims that ∆TUV is transformed to ∆WXY with a scale factor of 3/2. Is Theodore correct?
A. Yes, the triangles are similar with a scale factor of 3/2.
B. No, the triangles are similar with a scale factor of 2/1.
C. No, the triangles are similar with a scale factor of 2/3.
D. No, the triangles are similar with a scale factor of 4/3.
Answer:
C. No, the triangles are similar with a scale factor of 2/3.
Step-by-step explanation:
∆TUV is the original triangle. After transformation, the size reduced to give us ∆WXY. This means ∆TUV was reduced by a scale factor to give ∆WXY. The scale factor should be a fraction, suggesting, the original size of the ∆ was reduced upon transformation.
Thus, the ratio of their corresponding sides = the scale factor.
This is:
If you multiply the side length of ∆TUV by ⅔, you'd get side length of ∆WXY.
So, Theodore is wrong.