No the sum would always be 360 degrees
Answer:
There are 42 quarters (and 6 dimes).
Step-by-step explanation:
The given ratio is 2:14. But this does not tell us how many dimes nor how many quarters we have. Instead, it's a ratio. Modify this by multiplying numerator and denominator by n and then find n:
Number of dimes 2n
----------------------------- = -----
Number of quarters 14n
Then the total number of coins is 2n + 14n =48, implying that:
16n = 48, or n = 3.
Then the number of dimes is 2(n) = 2(3) = 6, and that of quarters is 14(n) =
14(3) = 42.
There are 42 quarters (and 6 dimes). Notice how the total of 42 and 6 is 48.
Answer:
-1 = 1/3 * 6 - 3
Step-by-step explanation:
So if it goes through point (6, -1) and has a slope of -1/3, all of this is parts of the slope intercept equal, Y = mx - B. You have the slope, which is m in the equation, y = -1/3 x - b. you already have a y and x which are the points passed, -1 = -1/3 * 6 - b, but this can't equal y, so you must have did it wrong, the only way this could equal y is if the slope wasn't negative, and in that case it would be -1 = 1/3 * 6 - 3. Because 1/3 * 6 is 2 and then subract 3 and you get -1.
Answer:
the least integer for n is 2
Step-by-step explanation:
We are given;
f(x) = ln(1+x)
centered at x=0
Pn(0.2)
Error < 0.01
We will use the format;
[[Max(f^(n+1) (c))]/(n + 1)!] × 0.2^(n+1) < 0.01
So;
f(x) = ln(1+x)
First derivative: f'(x) = 1/(x + 1) < 0! = 1
2nd derivative: f"(x) = -1/(x + 1)² < 1! = 1
3rd derivative: f"'(x) = 2/(x + 1)³ < 2! = 2
4th derivative: f""(x) = -6/(x + 1)⁴ < 3! = 6
This follows that;
Max|f^(n+1) (c)| < n!
Thus, error is;
(n!/(n + 1)!) × 0.2^(n + 1) < 0.01
This gives;
(1/(n + 1)) × 0.2^(n + 1) < 0.01
Let's try n = 1
(1/(1 + 1)) × 0.2^(1 + 1) = 0.02
This is greater than 0.01 and so it will not work.
Let's try n = 2
(1/(2 + 1)) × 0.2^(2 + 1) = 0.00267
This is less than 0.01.
So,the least integer for n is 2
