The equation for a circle is x^2 - 8x + y^2 -8 =0. what is the equation of the circle in standard form?
1 answer:
You have to complete the square on this to get it into standard form of a circle. Move the 8 over to the other side because that's part of the radius. Group together the x terms, take half the linear term which is 8, square it and add it in to both sides. Half of 8 is 4, 4 squared is 16, so add in 16 to both sides. I'll show you in a sec. You don't need to do anything to the y squared term. This just means that the center of the circle does not move up or down, only side to side, right or left. Here's your completing the square before we simplify it down to its perfect square binomial. 
. Now break down the parenthesis into the perfect square binomial and do the addition of the right: 
. This is the standard form of a circle that has a center of (4, 0) and a radius of 
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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>)
</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
Answer:
Step-by-step explanation:
We are given the following in the question:
The needle size should not be too big and too small.
The diameter of the needle should be 1.65 mm.
We design the null and the alternate hypothesis
Sample size, n = 35
Sample mean, = 1.64 mm
Sample standard deviation, s = 0.07 mm
Type I error:
- It is the error of rejecting the null hypothesis when it is true.
- It is also known as false positive error.
- It is the rejecting of a true null hypothesis.
Thus, type I error in this study would mean we reject the null hypothesis that the average diameter is 1.65 mm but actually the average diameters of the needle is 1.65 mm.
Thus, average diameter is 1.65 mm and we decide that it is not 1.65 mm.