A^2 + b^2 = c^2
3^2 + 8^2 = c^2
9 + 64 = c^2
73 =c^2
take the square root of each side
c= 8.544
3^2 + 8^2 = c^2
9 + 64 = c^2
73 =c^2
take the square root of each side
c= 8.544
While at the beach Daniel buys lunch for his family from a food stand he purchased one hotdog for $2.50 and three hamburgers if
1 answer:
The cost of each hamburger is $ 3.5
<em><u>Solution:</u></em>
Given that, Daniel purchased one hotdog for $2.50 and three hamburgers
He spent $ 13 in total
Let "H" be the cost of each hamburger
From given,
one hotdog for $2.50 + three hamburgers each for "H" = $ 13
Thus we get,
Thus cost of each hamburger is $ 3.5
You might be interested in
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.