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 is
sin (a+b)=-36/325

8 0

3 years ago

Which are the solutions to the quadratic equation x^2=9x+6

AlekseyPX

I will use a quadratic formula:

$x^2=9x+6\ \ \ \ |-9x-6\\\\x^2-9x-6=0\\\\a=1;\ b=-9;\ c=-6\\\\b^2-4ac=(-9)^2-4\cdot1\cdot(-6)=81+24=105\\\\\sqrt{b^2-4ac}=\sqrt{105}\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\x_1=\dfrac{-(-9)-\sqrt{105}}{2\cdot1}=\dfrac{9-\sqrt{105}}{2}\\\\x_2=\dfrac{-(-9)+\sqrt{105}}{2\cdot1}=\dfrac{9+\sqrt{105}}{2}$

5 0

3 years ago

What is the proportion of 2/5 = 8/x

natali 33 [55]

To find is out put it like this 2 8
5 x
you multiple so 2 times what eduals 8 then you take that and multiply by 5

4 0

3 years ago