How many solutions 4x2 plus 8x plus 3 equals 0?
Start with the equation 4x2 + 8x + 3 = 0.
Invoke the quadratic equation:
x = [-b ± √(b2 - 4ac)]/2a, where a = 4, b = 8, and c = 3.
Plug in the values for a, b, and c and solve:
x = [-8 ± √(82 - 4*4*3)]/2*4,
x = [-8 ± √(64 - 48)]/8,
x = -1 ± √(16)/8,
x = -1 ± 1/2,
x = {-1/2, -3/2}.
What is x in the equation 12 to the 2x power equals 28?
(12)2x = 28
(2x) log(12) = log(28)
2x = log(28) / log(12) = 1.34098 (rounded)
x = 0.67049 (rounded)
The lines are perpendicular.
What is 2X plus 3 equals 6 plus 4X?
First, let's put that information into an equation. 2x + 3 = 6 + 4x
Subtract 2x from both sides.
3 = 6 + 2x
Subtract 6 from both sides.
-3 = 2x
Divide each side by 2.
x = -3/2
How do you solve 3y-4j equals 9?
Express one variable in terms of the other.
3Y - 4J = 9
3Y = 4J + 9
Y = (4J + 9)/3
------------------------
insert back into original equation
3(4J + 9)/3 = 9
you see the 3 factors cancel
4J + 9 = 9
4J = 0
J = 0
-------------you have J
3Y - 4(0) = 9
3Y = 9
Y = 3
----------------one set of solutions to this problem
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Cameron Escalante is 19 years old (born September 19, 1997).
What is the differentiation of y with respect to x for x equals y exponent y?
x = yy differentiate both sides with respect to x
dx = (y * yy-1) dy
dy/dx = y * yy-1
dy/dx = yy = x
hence differentiate of y wrt x is x only
2x+4=8
2x=8-4 (you subtract 4 on both sides)
2x=4 (you then divide by 2 on both sides)
x=2
How do you derive the formula for the simple EOQ model?
The simple, or basic, economic order quantity (EOQ) is a special case of the continuous rate EOQ, which can be derived from the equation of total cost as follows.
Here is the equation for total cost (TC) as a function of run size (q):
TC(q) = K*D/q + P*D + q*H(r - D)/(2r), where:
K = Fixed cost per order
D = Annual Demand of product
q = run size
P = Purchasing cost per unit
H = Annual holding cost per unit
r = Production rate
K*D/q = Setup cost
P*D = Purchasing cost
H(r - D)/(2r) = holding cost.
To find the maximum value of q, you take the derivative, d[TC(q)]/dq, set it equal to zero, and solve for q.
First, take the derivative:
d[TC(q)]/dq = -K*D/q2 + H(r - D)/(2r).
Then, to maximize, set this equal to zero, and solve for q:
H(r - D)/(2r) - K*D/q2 = 0,
q2 = (2*r*K*D)/[H(r - D)],
q = √((2*r*K*D)/[H(r - D)]).
That's the formula for the continuous rate EOQ.
Basic EOQ is the special case of r >> D, which means r - D pretty much equals r, which allows you to cancel the r's in the above equation, giving you the formula:
q = √((2*K*D)/H).
This is the formula for basic EOQ.
What is x to the problem 68 plus 28x-28 equals 236?
68 + 28x - 28 = 236
28x + 68 - 68 -28 + 28 = 236 - 68 + 28
28x = 196
x = 7
Why do you need to find the inflection point on a graph?
To find the inflection points on a graph, you need to take the second derivative. Then, set that equal to zero to find the x value(s) of the inflection point(s).