What is the value of (j+3k)(5)? j(x) = (x + 4)^2; k(x) = 8 - x (j + 3k)(5) =
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Answer:
Step-by-step explanation:
x² + 2x - 3 + y² = 5
Strategy:
Convert the equation to the centre-radius form:
(x - h)² + (y - k)² = r²
The centre of the circle is at (h, k) and the radius is r .
Solution:
Move the number to the right-hand side.
x² + 2x + y² = 8
Complete the square for x
(Take half the coefficient of x, square it, and add to each side of the equation)
(x² + 2x + 1) + y² = 9
Complete the square for y
The coefficient of y is zero.
(x² + 2x + 1) + y² = 9
Express the result as the sum of squares
(x + 1)² + y² = 3²
h = -1; k = 0; r = 3
The centre of the circle is at
The graph of the circle below has its centre at (-1,0) and radius 3.
Given equation : .
Strategy 1: We can cross mutiply both sides remove fraction form.
On cross multiplication, we get
x * 7 = 3 * 42
7x = 126.
Dividing both sides by 7, we get
<h3>x = 18.</h3>Strategy 2: We can find least common denominator(lcd) of both sides and multiplying both sides by that lcd to get rid denominators from both sides.
LCD of 42 and 7 is 42.
Therefore, multiplying both sides by 42, we get
x = 6 * 3
<h3>x = 18.</h3>