F(\Xd7)=4\Xd7-5and G(\Xd7)=3\Xd7find (F+G(\Xd7)
F(×)=4×-5and g(×)=3×find (f+g(×)
"f(x) = 4x - 5, g(x) = 3x, find (f + g)(x)"
(f + g)(x) = f(x) + g(x)
(f + g)(x) = (4x - 5) + 3x
(f + g)(x) = 7x - 5
A Circle Having Radius 5 Units And Passing Through Point (4,3) And Its Centre Lies On Line 2x+Y-1=0.. Find The Equation Of Circle
A circle having radius 5 units and passing through point (4,3) and its centre lies on line 2x+y-1=0.. Find the equation of circle Answer: (x - 1)² + (y + 1)² = 5² or (x + 1)² + (y - 3)² = 5² Step-by-step explanation: We know that the equation of a circle is expressed as (x - h)² + (y - k)² = r² Given: r = 5 Unknown: (h, k) To solve for the center, were given the following clues; (1) the circle is passing through point (4, 3) (2) the center lies on the line 2x + y - 1 = 0 Identify the two equations in which you can get the center of the circle; Equation 1 = 2x + y - 1 = 0 Equation 2 = *I used the distance formula in equation 2. This will help me identify the missing center of the circle. Solve for y in equation 1: 2x + y - 1 = 0 2x + y = 1 y = 1 - 2x Substitute the value of y in equation 2: Substitute the value of x in equation 1: When x = 1, y = -1 When x = -1, y = 3 Substitute the values of x and y to (h, k). Yoll then get: (x - 1)² + (y + 1)² = 5² or (x + 1)² + (y - 3)² = 5² ^...
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