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- Question #ef67d - Socratic
What do you notice about these values? Try doing the x+y column minus the x column 20-4=16=y 20-3=17=y 20- (-40)=60=y 20-20=0=y Seems like we have a pattern From this, we can generalize with an equation to find the answer x+y=20 (by definition) Subtracting x from both sides y=20-x So for any x value, your y value is 20 minus your x
- How do you solve (5s + 7) ( s + 5) = 23? | Socratic
(5s +7)(s +5) = 23 Use BODMAS rule First solve brackets using distributive property: 5s(s + 5) + 7(s +5) = 23 5s2 + 25+ 7s +35 = 23 5s2 + 7s + 60 = 23 5s2 + 7s + 60 −23 = 0 5s2 + 7s + 60 −23 = 0 5s2 + 7s + 37 = 0 According to the Quadratic Formula, s , the solution for as2 + bs +c = 0 , where a, b and c are coefficients, is given by : s = − b ± √b2 − 4ac 2a In our case, a = 5 b = 7
- How do you evaluate \frac { 3} { x ^ { 2} - 9} + \frac - Socratic
We first try to get the fractions sorted (factorize) ->3 ( (x+3) (x-3))+11 (2 (x+3))=1 The LCM of the denominators is 2 (x+3) (x-3), so we multiply everything by that: (I'll spare you the in-betweens, but a lot cancels out): ->3xx2+11xx (x-3)=1xx2 (x+3) (x-3) Work out the brackets: ->6+11x-33=2x^2-18 Take everything to one side: 0=2x^2-11x+9
- How do you convert r=4 theta - sin theta +cos^2theta to . . . - Socratic
How do you convert r = 4θ − sin θ + cos2 θ to Cartesian form?
- How do you use the Change of Base Formula and a calculator . . . - Socratic
You should get 3: Actually using the definition of log you get: log_2 (8)=x 2^x=8 2^x=2^3 x=3 or: The Change of Base Formula tells you that: log_ab=log_cb log_ca where a is the old and c is the new base
- How do you solve #\frac { ( b + 1) } { 2} = \frac - Socratic
b=-3 Multiply both sides by 2 and both sides by 5 to get: 5 (b+1) = 2 (b-2) Then expand the brackets: 5b+5 = 2b-4 Then take all unknowns to one side, and all constant to the other to give: 5b - 2b = -4 -5 Then simplify: 3b = -9 Then divide both sides by 3: b = -9 3 The answer can then be simplified to: b = -3
- How do you use synthetic division to divide (14x^2 - Socratic
\\frac(14x^2-34)(x+4)=14x-56+\\frac{258}{x-4} The premise given factor (x-a) to f(x)=Bx^2+Cx+D, The pink arrow means multiply a by whatever the difference of the previous column's value was (it's subtraction) You start out with taking B and putting it in the difference row, so the next column will be C-aB The next column should be D-a(C-aB), and so on This applies to all polynomial
- How do you solve #\frac { 1} { 2} ( 2x + 4) ^ { 2} = 12#? - Socratic
Simplify #sqrt24# and factor #2x+4# so you can cancel out common factors Don't forget to add the #+-#! #2 (x+2)=+-2sqrt6# #x+2=+-sqrt6#
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