equations Flashcards
(34 cards)
If x³ +b³ = (x+b) (x²-xb + 9) and b>0 , what is the value of b
3
x³ (x² - 10) = -9x
if x>0, what is the solutions
c(ax−b)−d(b−ax)
Which of the following is equivalent to the expression above?
(ax-b)(c+d)
(ax-b)² (c+d)
(b-ax)²(c-d)
(ax-b)(b-ax)(c-d)
numberx^4 - number
This means, when an equation does’nt have a number with x , the factored form is always:
256a^4 - 1
(ax +b) (ax-b) (cx +b) ²
4x + 1) (4x - 1) (16x² +1
(y² - 32y +256 - k) (y² -32y+256 + k) = (y-16)^4 - 100
what is the value of k
(y² - 32y +256 ) = (y-16)²
Therefore, - k + k = 100
s³t³ + 1
a) (st-1) ( s²t²+st + 1)
b) (st- 1 ) (s²t² - st - 1)
c) (st-1) ( (st)²-st + 1)
c) (st+1) ( (st)²-st + 1)
1) Prove one equation and analyze how to get the correct answer
a) (st-1) ( s²t²+st + 1) : s³t³ +s²t² +st - s²t² - st - 1 -> s³t³ -1
“1” need to be +1 -> So, the last 1 should be both positives
c) (st+1) ( (st)²-st + 1)
w= 6+ 3xy/75
The equation above gives the quantity w in terms of the quantities x and y Which of the following equations correctly expresses yin terms of w and x?
a) y= 25w + 150 / x
b) y= 25w - 150 / x
c) ) y= 25w -2 / x
d) a) 25w + 2 / x
w= 6+ 3xy/75
(w - 6 ) 75 = 3xy
(75w - 75.6)/3 = xy
25w - 150) /x = y
y-b / x-a = m a= a) -mx + y - b /m b) mx + y - b /m c) y - b /m d) -y +b /m
- Eliminate the denominator first
y-b = m (x-a) -> y-b = mx - ma - Isolate a
y-b-mx/ m = -a - Multiplies -1 , so a be positive
(y-b+mx/ m) . -1 = a => - y-b+mx/ m
(1+1 /2) . -1 : the -1 multiplies only the fraction as a whole or also affects the sum inside “1+1”
a(1-x) + 5y = 18
MULTIPLIES THE FRACTION AS A WHOLE
18 - 5y = a(1-x)
18-5y/a - 1 = -x
(18-5y/a - 1). -1 = x => - 18-5y/a + 1
T= √(uqx)² + (To)²
Which of the following equations correctly gives x in terms of T, To, u?
a) x = T-To/uq
b) x = T-(To)²/uq
c) x= √T²-(To)²/uq
c) x= √T²-(To)²/uq
T= √(uqx)² + (To)² => T= (uqx)² + (To)²
√T - To² = √(uqx)²
v = 331.3 √1 + T/ 273.15
√ extends to the T fraction
(v / 331.3 - √1 )² = + (√T/ 273.15 )²
(v²/ 331.3² - 1 ) = + T/ 273.15
(v²/ 331.3² - 1 ) 273.15 = + T
(x-h)² + ( y - k)² = r²
a) y = r-x + h + k
b) y = r - (x-h)² + k
c) y = +- √r² - (x-h)² + k
r² - (x-h)² = (y-k)²
+ - √r² - (x-h)² = √(y-k)²
+ - √r² - (x-h)² + k = y
k² = m² + n²
a) m = k - n
b) m = √k-n
b) m = √k²-n²
b) m = √k-n
the square extends to k-n
The area, A, of the unshaded circular ring shown at left is given by the equation A= π(R²-r^²)
where R is the radius of the larger circle, and r is the radius of the smaller circle. Which of the following equations correctly gives R in terms of A and r?
a) R= √A / pi + r²
b) R= √A / pi + r² ( R= √extends to r²
c) R= √A / pi - r² ( R= √extends to r²
A/pi + r² = R²
√A/pi + r² =R B
n = (1.96o / E) ²
a) o = √En/1.96
B) o = E. √n/1.96
C) o = 1.96n /E
nE² /1.96²= o²
Simplify
√nE²/√1.96² = √o² -> √nE/1.96
√10X - 2xy = 4x
√ extends to -2xy
a) y = - 18x
b) y = -8x + 5
c) y = -2x + 5
d) -16x² - 5x
1) Eliminate the ratio
(√10X - 2xy)² = (4x )² -> 10x - 2xy = 16x²
2) Simplify by dividing 2
5x - xy = 8x²
3) Isolate the result
(8x² - 5x ) / - 1 x = -8x² + 5
let’s say we have A and B in the equation. But, there are two of B in the fraction position
R = C-P/ P
1) Eliminate the denominator by multiplying it both sides
RP = C- P
2) Again, eliminate the B ( who has a twin in the equation ) by adding B both sides
RP + P = C
3) Isolate the common : P ( 1+ R) = C
4) Isolate the desired one :P = C / (1+R)
let’s say we have A and B in the equation. But, there are two of B in the fraction position
C= ab / a + b
a =
1) Eliminate the denominator by multiplying both sides
C ( a+b) = ab -> Ca + Cb = ab
2) Eliminate the one that has twin and it’s going to be the result (f(x)
Cb = ab - Ca
3) Isolate the common
Cb = a (b-C)
4) Isolate the result
Cb/ b - C = a
1/f = 1/o + 1/i
o =
1/f - 1/i = 1/o ->
2) MMC
i - f /fi = 1/o
3) Eliminate the numerator 1 by inverting the ratio : potent ^-1
(i - f /fi)^-1= (1/o)^-1 => fi/ i - f = o
V = C ( 1-r)^t
r =
V/ C - 1 ^t = -r^t
(V/ C - 1 ). -1^t = r^t
RATIONALIZE THE ^t
^t√- V/ C + 1 = ^t√ r^t => (- V/ C)^1/t + 1 = r
L = Lo/x
x = 1/ √ 1 - v²/c²
Extends up to -v²/c²
v=
L = Lo/ 1/ √ 1 - v²/c²
INVERT 1/ √ 1 - v²/c²
L = Lo. √ 1 - v²/c²
L/Lo = √ 1 - v²/c²
ELIMINATE THE RATIO , by ²
T = √(ugx)²+ (To)²
√ extends to To)²
x=
√T² - To²/ug = x
√extends to To²
If f(x)=x²-6x and g(f(x))=-2x²+12x+5 what is the value of g(-9)?
1) f(x) = - 9 : x² - 6x = -9
x= 3
2) -2.3²+ 12.3 + 5 = -5 + 41 = 23
If f(x)= 7x -1 and f(g(x)) = 14x + 6, what equartion is equal g(x)
7 ( g(x) ) - 1 = f(g(x)) = 14x + 6
7. g(x) - 1 = 14x + 6
7. g(x) = 14x + 7
g(x) = 14x + 7 : 7 = 2x + 1
