X factor because its the u have to run before you fly effect
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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
So 20=2 itmes m2 times 5
5m=5 times m
so
5m=20
5 times m=2 times 2 times 5
divid eboth sides y 5
m=2 times 2
m=4
K = s + 4
k + s = 26
s + 4 = 26
Subtract 4 from both sides of the equation.
s = 22
Steve ran 22 miles.
Step-by-step explanation:
Notice: The literal factors are all the combinations of a and b where the sum of the exponents is 4: a4, a³b, a²b², ab³, b4 ... The solution to the problem of the binomial coefficients without actually ... The upper index n is the exponent of the expansion; the lower index k indicates which term ... 1a5 + a4b + a3b² + a²b3 + ab4 + b5.