A circle with centre (5,-4) passes through the point (5, 0). Find its equation.
x\(^2\) + y\(^2\) + 10x + 8y + 25 =0
x\(^2\) + y\(^2\) +10x - 8y - 25 = 0
x\(^2\) + y\(^2\) - 10x + 8y + 25 =0
x\(^2\) + y\(^2\) -10x - 8y - 25 = 0
Correct answer is C
(x - h)\(^2\) + (y - k)\(^2\) = r\(^2\)
x\(^2\) - 2hx + y\(^2\) - 2ky + h\(^2\) + k\(^2\) = r\(^2\)
x\(^2\) - 2(3)x + y\(^2\) - 2(-4) y + 5\(^2\) + (-4)\(^2\) = r\(^2\)
x\(^2\) - 10x + y\(^2\) + 8y + 25 + 16 = r\(^2\)
x\(^2\) - 10x + y\(^2\) + 8y + 41 = r\(^2\)
at point (5,0)
5\(^2\) - 10(5) + 0\(^2\) + 8(0) + 41 = r\(^2\)
25 - 50 + 41 = r\(^2\)
16 = r\(^2\)
r = \(\sqrt{16}\)
= 4
x\(^2\) + y\(^2\) - 10x + 8y + 25 = 0
Find the nth term of the linear sequence (A.P) (5y + 1), ( 2y + 1), (1- y),...
(8 + 3n)y + 1
8y + 3n + 1
(8 - 3n)y + 1
8y - 3n + 1
Correct answer is C
Tn = a + (n - 1)d
Tn = 5y + 1 (n - 1) -3y
Tn = 5y + 1 - 3ny + 3y
Tn = 8y - 3ny + 1
Tn = (8 - 3n) y + 1
If the binomial expansion of (1 + 3x)\(^6\) is used to evaluate (0.97)\(^6\), find the value of x.
0.03
0.01
-0.01
-0.03
Correct answer is C
(1 + 3x )\(^6\) = (0.97)\(^6\)
1 + 3x = 0.97
3x = 0.97.1
\(\frac{3x}{3} = \frac{0.03}{3}\)
x = -0.01
A sale of goods to audu was not posted. This is an error of
Omission
Compensation
Commission
Principle
Correct answer is A
No explanation has been provided for this answer.
N14,480
N12,000
N9,830
N7,130
Correct answer is C
No explanation has been provided for this answer.