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Note: Conversion is based on the latest values and formulas.
How do you simplify #\frac { \sqrt { 18x y - Socratic 3 Nov 2017 · Since 18 is the product of the perfect square 9 and 2, we can rewrite the radical as sqrt(9*2), which is equivalent to sqrt9sqrt2. Since 9 is 3^2, we can simplify sqrt9 to 3; we can't …
How do you find the distance between the points (-3,8), (5,4)? 19 Feb 2017 · The distance between the points is #sqrt(65)# or #8.062# rounded to the nearest thousandth. Explanation: The formula for calculating the distance between two points is:
(sqrt(49+20sqrt6))^(sqrt(asqrt(asqrt(a...oo)+(5-2sqrt6)^(x^2+x-3 22 Nov 2017 · #(sqrt(49+20sqrt6))^(sqrt(asqrt(asqrt(a...oo)#+#(5-2sqrt6)^(x^2+x-3 - sqrt(xsqrt(xsqrt(x...oo))))=10# where #a=x^2-3#, then x is?
How do you solve 2x^2 = 14x + 20? - Socratic 20 Apr 2018 · x = (7 +- sqrt(89))/2 => 2x^2 = 14x + 20 Divide both sides by 2 => x^2 = 7x + 10 Rearrange the equation => x^2 - 7x - 10 = 0 It’s in the form of ax^2 + bx + c = 0 ...
How do you solve 2x^2 + 7x = 5? - Socratic 31 Jan 2017 · x=(-7+sqrt89)/4; x=(-7-sqrt89)/4 Solve 2x^2+7x=5 Subtract 5 from both sides. 2x^2+7x-5=0 This is a quadratic equation in the form ax^2+7x-5, where a=2, b=7, c=-5.
What is the length o segment CD in the figure below? - Socratic 20 Feb 2018 · CD=8.06 The coordinates of C are (-3,1) and those of D are (4,5). Hence the base of right angled triangle formed in figure is 4-(-3)=4+3=7 units and height of right angled triangle …
If cos theta = 4/7 where theta is an acute angle, then what are the ... 22 Oct 2017 · sin theta = sqrt(33)/7 tan theta = sqrt(33)/4 csc theta = (7sqrt(33))/33 sec theta = 7/4 cot theta = (4sqrt(33))/33 SInce we are told that theta is an acute angle in a right triangle, we …
Why does sqrt(49) have two answers? - Socratic This is because there are two values that can be multiplied by themselves to produce 49 - one positive and the other negative. This is always the case for a square root. The positive value of …
How do you solve ln(x - 3) + ln(x + 4) = 1? - Socratic 14 Jul 2015 · x~=3.37 First thing to note is that a ln(x-3) is defined only when x-3>0=>x>3 And also ln(x+4) is defined when x+4>0 => x> -4 And so for both functions to be defined we need …
How do you solve #\sqrt { 2x + 5} - \sqrt { x + 14} = 1#? 5 Dec 2016 · x=22 Given: sqrt(2x+5) - sqrt(x+14) = 1 Aside: Note that x=2 would give -1 rather than 1, so we may encounter it as an extraneous solution later. Let us try adding sqrt(x+14) to …