The answer to what is 3/20 of a circle is 54 because we know a circle is 360 degrees so we divide that by 30 and we get 18. Now that we have 18 we multiply that by 3 and we get our answer of 54
Answer:
A
Step-by-step explanation:
Perpendicular bisector of a line divides the line into 2 equal parts and it is perpendicular to the line.
First let's find the midpoint of CD. The point is where the perpendicular bisector will cut through the line.
midpoint= 
Thus, midpoint of CD
Gradient of line CD
The product of the gradients of perpendicular lines is -1.
gradient if perpendicular bisector(1)= -1
gradient of perpendicular bisector= -1
y=mx +c, where m is the gradient and c is the y-intercept.
y= -x +c
Subst a coordinate to find c.
<em>Since the perpendicular bisector passes through the point (8, -10):</em>
When x=8, y= -10,
-10= -8 +c
c= -10 +8
c= -2
Thus, the equation of the perpendicular bisector is y= -x -2.
Answer:
23.5 is the answer
Step-by-step explanation:
X10 =235 so x=235÷10 = 23.5
<span>Multiply one of the equations so that both equations share a common complementary coefficient.
In order to solve using the elimination method, you need to have a matching coefficient that will cancel out a variable when you add the equations together. For the 2 equations given, you have a huge number of choices. I'll just mention a few of them.
You can multiply the 1st equation by -2/5 to allow cancelling the a term.
You can multiply the 1st equation by 5/3 to allow cancelling the b term.
You can multiply the 2nd equation by -2.5 to allow cancelling the a term.
You can multiply the 2nd equation by 3/5 to allow cancelling the b term.
You can even multiply both equations.
For instance, multiply the 1st equation by 5 and the second by 3. And in fact, let's do that.
5a + 3b = –9
2a – 5b = –16
5*(5a + 3b = -9) = 25a + 15b = -45
3*(2a - 5b = -16) = 6a - 15b = -48
Then add the equations
25a + 15b = -45
6a - 15b = -48
=
31a = -93
a = -3
And then plug in the discovered value of a into one of the original equations and solve for b.</span>
