in standard form find the center and radius of the circle.show all of your work for full credit
The given equation can be rewritten into the form of an equation for a circle with center (h,k) and radius r. After completing the squares for x and y coefficients, we determine the center as (2, -5/2) and radius as 1.
The provided equation is 2x2 + 2y2 - 8x + 10y + 2 = 0.Let's rewrite it in the form of an equation for a circle, which is (x-h)^2 + (y-k)^2 = r^2, where (h,k) represents the center of the circle and r represents the radius:
First, we group the terms related to x and y separately like this: 2(x2- 4x) + 2(y2 + 5y) +2 = 0.
Now, re-arrange the terms to complete the square: 2[(x-2)^2 - 4] + 2[(y+5/2)^2 - (5/2)^2] + 2 = 0.
Re-arrange again, this time to make it equate to zero: (x-2)^2 + (y+5/2)^2 = 1. Hence, the circle's center (h,k) is at (2, -5/2) and the radius (r) is the square root of 1, which is 1.
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Answer:
first group x's with x's and y's with y's
then complete the squra with x's and y's
2x^2-8x+2y^2+10y+2=0
2(x^2-4x)+2(y^2+5y)+2=0
take 1/2 of linear coeficient and square
-4/2=-2, (-2)^2=4
5/2=2.5, 2.5^2=6.25
add that and negative inside
2(x^2-4x+4-4)+2(y^2+5y+6.25-6.25)+2=0
factor perfect squares
2((x-2)^2-4)+2((y+2.5)^2-6.25)+2=0
distribute
2(x-2)^2-8+2(y+2.5)^2-12.5+2=0
2(x-2)^2+2(y+2.5)^2-18.5=0
add 18.5 both sides
2(x-2)^2+2(y+2.5)^2=18.5
divide both sides by 2
(x-2)^2+(y+2.5)^2=9.25
that is a circle center (2,-2.5) with radius √9.25
B. There is exactly one triangle because b sin A = 0.5b.
C. There are two triangles because b sin A < a < b.
D. There are three triangles because b sin A = 0.5b and b sin A < a < b.
Answer:C
Step-by-step explanation:
there are two triangles as
a=2
b=3
Using sine rule
sinB=0.75
B=131.41
For these two values two values of C is possible
calories. How many calories does the
smaller candy bar have?