Math Pre-cal Question

Pre-cal math problem?!?

i need help with the problem!! PLEASE!

A company produces a product for which the variable cost is $5.35 per unit and the fixed costs are $16,000. The company sells the product for $8.20 and can sell all that it produces.

a) find the total cost as a function of x, the number of units produced.

b) find the profit as a function of x

Trig/Pre-Cal Math Question?

Um, I can't remember how to do this problem...

Find a formula for An (as in "a of n") for the arithmetic sequence.
a7 = 8
a13 = 6

The formula I am trying to find is along the lines of an = d(n-1) + a1

Precal polynomial question?

i have some true or false questions from my precal class.
1) if 2-3i is a zero of a polynomial with real coefficients, then so is -2+3i

2)if the determinant of th coefficient matrix is 0 the corresponding system has no solution

3) most trigonometric equation have unique solutions

4) the graphs of y=tanx, y=cotx, y=secx, and y=cscx each have infinitely many vertical asymptotes

if any one can help me, i will very apperciated!

Simple math precalculus question factor sec^3x - sec^2x - secx + 1?

factor sec^3x - sec^2x - secx + 1

and while we are n the subject
how do I verify these algebraically

(secx-tanx)(cscx+1)=cotx

1+cscx/cotx+cosx=secx

sinx+cosx/sinx - cosx-sinx/cosx=secxcscx

Please do not just simply give me the answer but explain how it is is reached.

Math pre-cal word problem?

the equation should be 2 = (1.029)^t

take the ln of both sides
ln2 = tln(1.029)
ln2/ln(1.029) = t
24.24655 = t

Math (Precal) Proving Identities?

Given:
2cot(2y) = (cot^3(y)-tan^3(y))/(sec^2(y)+cot^2(y))

Rewrite using sin and cos:
2cos(2y)/sin(2y) =
(cos^3(y)/sin^3(y)-sin^3(y)/cos^3(y)) /
(1/cos^2(y)+cos^2(y)/sin^2(y))

Apply double angle formulas:
2(cos^2(y) - sin^2(y))/(2sin(y)cos(y)) =
(cos^3(y)/sin^3(y)-sin^3(y)/cos^3(y)) /
(1/cos^2(y)+cos^2(y)/sin^2(y))

Multiply both sides by sin*cos:
2(cos^2(y) - sin^2(y))/2 =
(cos^4(y)/sin^2(y) - sin^4(y)/cos^2(y)) /
(1/cos^2(y) + cos^2(y)/sin^2(y))

Multiply top and bottom of right side by sin^2cos^2
cos^2(y) - sin^2(y) =
(cos^6(y) - sin^6(y) / (sin^2(y) + cos^4(y))

Multiply both sides by (sin^2(y) + cos^4(y))
(cos^2(y) - sin^2(y))(sin^2(y) + cos^4(y)) =
(cos^6(y) - sin^6(y)

Multiply terms:
(sin^2(y)cos^2(y) - sin^4(y) + cos^6(y) - sin^2(y)cos^4(y) =
(cos^6(y) - sin^6(y)

Subtract cos^6(y) from both sides:
(sin^2(y)cos^2(y) - sin^4(y) - sin^2(y)cos^4(y) = - sin^6(y)

Divide both sides by sin^2(y):
(cos^2(y) - sin^2(y) - cos^4(y) = -sin^4(y)

Rearrange and factor:
(cos^2(y) - cos^4(y) = sin^2(y) - sin^4(y)
cos^2(y) * [(1 - cos^2(y)] = sin^2(y) * [1 - sin^2(y)]

Apply identity cos^2 + sin^2 = 1
cos^2(y) * [sin^2(y)] = sin^2(y) * [cos^2(y)]

Re-order:
sin^2(y)cos^2(y) = sin^2(y)cos^2(y)

Since both sides are identical, the identity is true.


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Simple Math problem precal?

Form equations from the given information

Angelo's age is 4 years more than twice Carl's age.

1) a = 2c + 4

Brandon is 5 years younger than Carl.

2) b = c - 5

The Average of the tree ages is 41

3) (a + b + c)/3 = 41

a + c + c = 123

You have a system of 3 equations in 3 unknowns

(2c + 4) + (c - 5) + c = 123 {substitute for a from equation 1 and for b from equation 2}

2c + 4 + c - 5 + c = 123

4c = 124

c = 31

1) a = 2(31) + 4 = 62 + 4 = 66

2) b = 31 - 5 = 26

check

(66 + 26 + 31)/3 = 41

123/3 = 41

41 = 41

Math problem.. Investment doubles.. pre-Calculus ?

Continuous compounding within the precision specified doesn't help all that much over annual compounding or other rates.

First, using the rather crude "Rule of 72",
we divide 72/13 and get 5.54 years as a rough estimate.

Or we can solve 1.13^x = 2
x log 1.13 = log 2
x = log 2 / log 1.13 = 5.67141717 with annual compounding

(Rule of 72 not doing badly, I'd say.)

Semi-annual compounding:
1.065^x = 2, x = 11.006739 half years or 5.50 years
Quarterly: 1.0325^x = 2, x = 21.6723318 = 5.42 years
13 times/year: 1.01^x = 2, x = 5.36 years
26 times/year: 1.005^x = 2, x = 5.34522006 years
52 times/year: 1.0025^x = 2, x = 5.33856349 years
104 times/year: 1.00125^x = 2, x = 5.33523313 years
208 times/year: 1.000625^x = 2, x = 5.33356743 years

I don't know how to take the limit, but I'll guess it's 5 1/3 years.

And all this over the "Rule of 72"
improved the accuracy by about 1/4 year.