Note that for x > 1 the numerator will be positive and the denominator negative so x/ (1-x) < 0 and the square root is undefined (at least not Real). So we can eliminate options (a) and (b).

The function is defined here. graph {y=sqrt ( (x-3)/x) [-12.66, 12.65, -6.33, 6.33]} In the question you are asking why the asymptote is x=0. Remember that an asymptote is a line that a curve approaches, it …

Mar 12, 2017 · Isolate the square root by subtracting 10 from both sides. √6x − 5+10 −10 = 3 −10 ⇒ √6x −5 = −7 square both sides (√6x − 5)2 = (− 7)2 ⇒ 6x − 5 = 49 add 5 to both sides. 6x−5 +5 = 49+ 5 ⇒ …

How do you simplify 4 times the square root of 8 over the square root of 2? How do you graph #sqrtx", if "x>0 and -1/2x+3", if "x≤ 0#? What is the equation for the circle which passes through (3;4) (-4;3) …

If you were asked to find the square root of 117 then you would first find: 117 = 3*3*13 So sqrt (117) = 3*sqrt (13) What is sqrt (13)?

#x+sqrt (x^2+4)=color (red) (2y)color (white) ("xxxxxx")#...or at least that's what I got

Jun 25, 2018 · In your case, the function y=sqrt (2x-7) has some restriction: you can't give any number as input, since a square root only accepts non-negative numbers. For example, if you choose x=1, …

No Counter example: color (white) ("XXXXXX")sqrt (1/4) = 1/2 but 1/2 is not less than the original number, 1/4. However, since I like Maggie, I will concede that her statement is true for all numbers …

x=5," extraneous solution " x=2 color (blue)"square both sides" (x-3)^2= (sqrt (x-1))^2 rArrx^2-6x+9=x-1 Equate the quadratic to zero rArrx^2-7x+10=0 Factorising the quadratic gives. (x-2) (x-5)=0 rArrx=5" …

Square the number and then multiply the answer by 4. This will produce corresponding y values. The quadratic is positive so the parabola is uu shaped. [-b]/ [2a]=0/8=0 this is the x coordinate of the …