|2x - 5/3| > 1 ???
This is what im getting but dont know if its correct?
.....set up 2 equations ..one positive..one negitive
|2x - 5/3| > 1 ............. |2x - 5/3|
multiply both sidesby 3 to rem
From a deck of 52 cards, the 12 face cards and 4 aces are removed. From these 16 cards that are removed, 4 are chosen. How many combinations are possible that have at most 1 red card?-Of the 16 cards
I have two points on this graph. One is (1,0), the other is (0,∞). It looks like half a parabola or hyperbola.
The x values range does not need to be just 1 to 0 - it may be drawn as 100 down to 0, bu
ok so my tmobile bill is $450 i ask my mom for $250 and my father for $250 i go pay my bill and i have $50 extra so i give my mom $10 and my dad $10 i keep $30 so my question is if you add $240 plus $
please help I have a test tomorrow and I need someone to explain this to me please!!!!!!!!!!!!!
In triangle ABC, D is the midpoint of AB and E is the midpoint of AC. If BC=7x+1, DE= 4x-2 and the meas
I need someone to solve the following problem.Please show and explain the steps to get the answer.
5-3(x-7(x-6)
The value of x is not given so just write it in standard form.-5-3(x-7(x-6))
=5-3(x-7x
Use Riemann sums and limits to compute the area bounded by
f(x)=15 x^3 + 6 x
and the x axis between x=2 to x =8
Thank you!!-dx = (6/n)
x(i) = 2+(6/n)i
i = 0, n-->inf
area = sum [i = 0, n] f(x(i)) dx
solve the following system of equations by the addition or subtraction method.
7x+y=13
y=2x-0.5
solve the following systems of equations using the substitution method:
x+5y=4
x=5y-16
thank you for h
1.8
2.+or-5
3.+or-10
4.+or-6-Any tangent to the ellipse x^2/9+y^2/16=1 is
y=mx(+or-)sqrt(9m^2+16) [y=mx(+or-)sqrt(a^2m^2+b^2)]
when m=-1 (slope of given line=-1)
the equation becomes
y=-x(+or-)sqrt(9+
Find the inverse laplace of
e^(-10s)/s(s^2+2s+2)-Ignore the exponential for a moment and do a partial fraction decomp on the rest.
1/[s(s² + 2s + 2)] = A/s + (Bs + C)/(s² + 2s + 2) ==> A =1/2 , B
plzz, i need help. i am going to select best answer.......-Suppose to the contrary that
m^2 = 1! + 2! + 3! + 4! + ... + n! for some integer m.
Reducing mod 5 yields
m^2 = 1! + 2! + 3! + 4! + 0 (mod 5
First, get a common denominator.When you factor x^2-4, you get (x+2) and (x-2), so you can multiply everything by these variables.
3(x+2)/((x-2)(x+2)) + 8/(x^2-4) = (x^2-4)/(x^2-4)
Since the denominat
Use the given values of n and p to find the minimum usual value μ - 2σ and the maximum usual value μ + 2σ. Round
your answer to the nearest hundredth unless otherwise noted.n = 194, p = 0.16
the answ
In quadrilateral ABCD, BC is parallel to AD. E is the foot of the perpendicular from B to AD. Find BE, if AB = 17cm, BC = 16cm, CD = 25cm and AD = 44cm.-Draw a rough sketch to visualize the problem. D