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Otrada [13]
3 years ago
13

Explain why sine of an angle greater than 180° and less than 360° is negative.

Mathematics
1 answer:
exis [7]3 years ago
8 0

9514 1404 393

Explanation:

The sine of an angle in standard position is the y-coordinate of the intersection of its terminal ray with the unit circle. When the angle is in the range 180° < θ < 360°, that point of intersection is in quadrant III or IV, where y-coordinates are negative.

__

<em>Alternative explanation</em>:

That is where the sine curve is below the x-axis.

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dybincka [34]

Answer:

121/9

Step-by-step explanation:

= (-3\frac{2}{3} )^2

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= \frac{121}{9}

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3 years ago
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Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)
DENIUS [597]

Compute the necessary values/derivatives of f(x) at x=1:

f(1)=3

f'(1)=2

f''(1)=-12

f'''(1)=-12

f^{(4)}(1)=48

f^{(5)}(1)=120

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) f(x) at x=1 by

T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5

T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5

###

Another way of doing this would be to solve for the coefficients a,b,c,d,e,g in

f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5

by expanding the right hand side and matching up terms with the same power of x.

5 0
3 years ago
Jason's mom drove at 100km/hr for a 160km trip.how much longer would the trip have taken if she had driven at 90km/hr
kondor19780726 [428]
We use the formule: t=S/v. So:
+ t1= 160/100 = 8/5 hrs = 5760''
+ t2= 160/90 = 16/9 hrs = 6400''

=> t2-t1= 6400 - 5760 = 640'' = 10'40''


3 0
3 years ago
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