Enter the opposite of 4/5 and − 7/15.
2 answers:
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Answer
w=7.6
x=43.7
Step-by-step explanation:
To solve for w you can use Sin(x)= which would basically be Sin(50)=
then solve. Multiply the ten on both sides so you have 10*Sin(50)=w and your final answer is 7.6
To solve for x you can also use Sin(x)= then just use Inverse of Sin^-1
The answer is x=43.7
Answer:
Question A)
Question B)
Question C)
Step-by-step explanation:
A)
We are given:
We can rewrite the expression:
The square root and square will cancel each other out. Thus:
B)
We are given:
Rewrite:
Note that the square root of 9 is simply 3. We can also factor the second part:
Rewriting:
Simplify:
C)
We are given:
Rewrite. Note that 50 = 25(2):
Simplify. We can rewrite the factor as:
The square and square root will cancel out. Thus:
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>)
</span>
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 is
sin (a+b)=-36/325
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>)
</span>
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 is
sin (a+b)=-36/325