Answer:
-3/5 as a decimal iss -0.6
Step-by-step explanation:
The complete question in the attached figure
we have that
tan a=7/24 a----> III quadrant
cos b=-12/13 b----> II quadrant
sin (a+b)=?
we know that
sin(a + b) = sin(a)cos(b) + cos(a)sin(b<span>)
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step 1
find sin b
sin²b+cos²b=1------> sin²b=1-cos²b----> 1-(144/169)---> 25/169
sin b=5/13------> is positive because b belong to the II quadrant
step 2
Find sin a and cos a
tan a=7/24
tan a=sin a /cos a-------> sin a=tan a*cos a-----> sin a=(7/24)*cos a
sin a=(7/24)*cos a------> sin²a=(49/576)*cos²a-----> equation 1
sin²a=1-cos²a------> equation 2
equals 1 and 2
(49/576)*cos²a=1-cos²a---> cos²a*[1+(49/576)]=1----> cos²a*[625/576]=1
cos²a=576/625------> cos a=-24/25----> is negative because a belong to III quadrant
cos a=-24/25
sin²a=1-cos²a-----> 1-(576/625)----> sin²a=49/625
sin a=-7/25-----> is negative because a belong to III quadrant
step 3
find sin (a+b)
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
sin a=-7/25
cos a=-24/25
sin b=5/13
cos b=-12/13
so
sin (a+b)=[-7/25]*[-12/13]+[-24/25]*[5/13]----> [84/325]+[-120/325]
sin (a+b)=-36/325
the answer issin (a+b)=-36/325
Let p (x) = 4x^4-13x^2-2x
The zero of x-2 is 2
putting x = 2 in p (x), we get,
p (2) = 4×2^4-13×2^2-2×2
= 64 - 52 - 4
= 64 - 56
= 8
Therefore, remainder = 8
Answer:
y=12/5 (in decimal notation y=2.4)
Step-by-step explanation:
the value of 7y-2 = 10 more than 2y
7y-2 = 10+2y
7y-2y = 10+2
5y = 12
y = 12/5
y = 2.4
Answer with Step-by-step explanation:
We are given that a function f(x) is continuous on (
).
1.f'(-1)=0 and f''(-1)=-7
We have to find information about f.
When f'(-1)=0 and f''(-1)=-7 < 0
Then, function is maximum at x=-1.
Therefore, at x=-1, f has local maximum.
Answer:a)at x=-1 ,f has local maximum.
2.) if f'(4)=0 and f''(4)=0
We know that when f''(x)=0 then test fails then the function has not maximum or minimum.
Therefore, at x=4 , f has not a maximum or minimum.
Answer:c) at x=4, f has not a maximum or minimum.