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## Answers by Priti S.

8 ≤ 3/4(2x+5) ≤ 6      avoid working with fraction so multiply the whole equation by 4 32 ≤ 3(2x+5) ≤ 24    Now distribute 3 32 ≤ 6x+15 ≤ 24      subtract 15 to front and back 32-15 ≤ 6x ≤ 24-15  simplify 17 ≤ 6x ≤ 9  ...

ax2+bx+c = 0, Quadratic formula:  x = [ -b ± sqrt(b^2 - 4ac) ] / 2a     1.   x^2-8x+20=0 a=1, b=-8 , c=20  Plug in the value for a, b and c in the quadratic formula     x = [ -b ± sqrt(b^2 - 4ac) ] / 2a x =...

5(t+3)+9 = 3(t-2)+6   Use distribution property  5t+15 + 9 = 3t-6 + 6 Combine like terms on same sides 5t+ 24 = 3t                Now combine like terms by subtracting 5t on both side -5t  ...

|x|+2 < 13             To solve absolute value problems, absolute value has to be all by itself.    Therefore move over the 2 by subtracting it to the other side of equal sign.   |x|+2  -2  < 13  -2  ...

Given: 2x-y=4 x+5y=24      Using substitution Solve for y in 1st equation 2x-y = 4     (add y over and subtract 4 over to other side to make y all by itself) y = 2x- 4   Now take 2nd equations and substitute for y  since...

Start out writing equation in Slope Intercept form: Y= MX + B   Use the point and slope to plug into the equation so you can solve for B. (1,2) and Slope (M) = 7   2 = (7)(1) + B 2 = 7 + B (Subtract 7 to find B) B = -5   Now write...

Use the distance formula: √( (x2-x1)² + (y2-y1)²)   Given : (x1,y1) & (x2,y2) For ex: (2,7) and (5,11)   =√( (5-2)² + (11-7)²) =√( (3)² + (4)²) =√(9 + 16) =√(25) = 5

9m^3n-36mn^3+39m^4n^2  Find the GCF (Greatest Common Factor) : 3 m n   3 m n (3 (m^2)-12 (n^2) + 13 (m^3) n)

If the equation is x^2 - 18x + 79 = 0   Completing the Square : move the number 79 on the other side of equal sign and use formula as indicated x^2 +bx +(b/2)^2  = c + (b/2)^2   x^2 - 18x  (-18/2)^2 = -70 + (-18/2)^2 x^2 - 18x...

a.  Use substitution to solve   (substitute for y and set them equal to each other) y = 5x - 3 y = 3x - 1    5x - 3 = 3x - 1     -3x       -3x ______________ 2x - 3 = - 1     +3     +3 ______________ 2x...

5√75 -4√50 + √48 = take this equation and break them into perfect square factors. 5√(25*3) -4√(25*2) + √(16*3)  Perfect square can come out see below for the rules 5*5√3 -4*5√2 + 4√3  25√3 -20√2 + 4√3   Now you can old add / subtract the front...

Start out with distributing √2 to the bracket(√2+y√6) √2*√2+y√2*√6 √4+y√12  Now simplify by bringing out the perfect square numbers. 2+y*2√3 2+2*y√3   I hope this helps   Memorize this and it really get easy 1^2 = 1  &...

what you have found is correct but i think you accidentally lost your sign   -12/8 is correct which will simplify as -3/2  or -1.5 .   Next you will have to write out the slope intercept form equation   Y = MX + b   Now plug in the order pair...

Write this equation in Slop-intercept form and it would be the best way.   Y= MX + b    Where M => Slope and b => y-intercept and (0,2) is already a y-intercept Y=-3x/5 + 2   I hope this helps!!

the conversion from decimal to minuets is 60. 1 hours has 60 mins.   so you take the .8 *60 = 48 mins so total 24 hrs and 48 mins.     for example 1.5 hours so take .5*60 = 30 mins   I hope this helps!

For factoring start out with (  ) (  ) double brackets   factor the first term 5x^2 (which is 5x & 1x place them as indicated below) (5x     )(x    )   Now think of factor of -72  For example (8 & 9 , 12 & 6, 1 & 72,...

rewrie it as you can see the form   (x^2-5)^2   now write them twice  (x^2-5) (x^2-5) FOIL x^4 -5x^2 -5x^2 +25 combine like terms x^4 -10x^2+25

Hello   Well you are not getting the correct equation   x+3y=15 3y=-x+15 now divide 3 to both x and 15 y= -x/3 +15/3   so y=-x/3+ 5 now pick points for x (i pick the easy one to solve for) and solve for y   x...

The vertex form of a parabola's equation is generally expressed as :    y=a(x−h)^2+k Where (h,k) is the max or min.   1.      Look at the graph.   2.      Find the vertex of the graph (h,k) 3.      Vertex...

Try this link and it has given all the different sample problems with explanation.  There are at least 7 problems with different scenario. http://hotmath.com/hotmath_help/topics/simplifying-radical-expressions.html I hope this helps.

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