What is 3.2 times 0.3 equils
2 answers:
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The answer is 42.
The question is asking for angle N. We know from the question that angle N is equal to angle E (Don't think about the scale factor because it only applies to the sides. They're just trying to trick you).
To find angle E:
Move numbers to the right side:
Combine like terms:
Divide both sides by 13:
The formula for angle E:
Plug in the 8:
The angle measure of E is 42. So angle N is automatically 42 too.
To solidify, we can try to use the formula which they give us for angle N:
The question is asking for angle N. We know from the question that angle N is equal to angle E (Don't think about the scale factor because it only applies to the sides. They're just trying to trick you).
To find angle E:
Move numbers to the right side:
Combine like terms:
Divide both sides by 13:
The formula for angle E:
Plug in the 8:
The angle measure of E is 42. So angle N is automatically 42 too.
To solidify, we can try to use the formula which they give us for angle N:
Answer:
The values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.
Step-by-step explanation:
Note: This question is not complete. The complete question is therefore provided before answering the question as follows:
Determine the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001. f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0
The explanation of the answer is now provided as follows:
Given:
f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0 …………….. (1)
= (x) = (e^z /4!)x^4
Since the aim is (x) < 0.001, this implies that:
(e^z /4!)x^4 < 0.0001 ………………………………….. (2)
Multiply both sided of equation (2) by (1), we have:
e^4x^4 < 0.024 ……………………….......……………. (4)
Taking 4th root of both sided of equation (4), we have:
|xe^(z/4) < 0.3936 ……………………..........…………(5)
Dividing both sides of equation (5) by e^(z/4) gives us:
|x| < 0.3936 / e^(z/4) ……………….................…… (6)
In equation (6), when z > 0, e^(z/4) > 1. Therefore, we have:
|x| < 0.3936 -----> 0 < x < 0.3936
Therefore, the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.