We know that
sin²x+cos²x=1
so
clear cos x
cos x=(+/-)√[1-sin²x]
in this problem
<span>Angle 0 is in quadrant 1 -----> cos o and sin o are positive
</span>sin o=2/5
cos x=√[1-(2/5)²]----> cos o=√[1-4/25]----> cos o=√[21/25]---> cos o=√21/5
the answer is
cos o=√21/5
Answer:
Σ(-1)^kx^k for k = 0 to n
Step-by-step explanation:
The nth Maclaurin polynomials for f to be
Pn(x) = f(0) + f'(0)x + f''(0)x²/2! + f"'(0)x³/3! +. ......
The given function is.
f(x) = 1/(1+x)
Differentiate four times with respect to x
f(x) = 1/(1+x)
f'(x) = -1/(1+x)²
f''(x) = 2/(1+x)³
f'''(x) = -6/(1+x)⁴
f''''(x) = 24/(1+x)^5
To calculate with a coefficient of 1
f(0) = 1
f'(0) = -1
f''(0) = 2
f'''(0) = -6
f''''(0) = 24
Findinf Pn(x) for n = 0 to 4.
Po(x) = 1
P1(x) = 1 - x
P2(x) = 1 - x + x²
P3(x) = 1 - x+ x² - x³
P4(x) = 1 - x+ x² - x³+ x⁴
Hence, the nth Maclaurin polynomials is
1 - x+ x² - x³+ x⁴ +.......+(-1)^nx^n
= Σ(-1)^kx^k for k = 0 to n
Differentiating an integral removes the integral.
f(x) = integral of dt/sqrt(t^3 + 2)
f'(x) = 1/sqrt(x^3 + 2)
f'(1) = 1/sqrt(1^3 + 2)
f'(1) = 1/sqrt(3) = sqrt(3)/3.
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Answer:
30 days
Step-by-step explanation:
The common multiples of 15 and 6 are 30, 60, 90 etc and because it increases by 30 each time we can find out that he will play both football and basketball together in thirty days!
