6 ÷ 10 I need help show my work and knowing the answer

cuanto dinero se debe invertir a un interes de 14% compuesto continuamente para tener $500,000 en 35 años?

Alja [10]

Answer:

Se necesita depositar un capital inicial de $ 5096.83 pesos para lograr $ 500000 pesos a un interés compuesto de 14 % anual luego de 35 años.

Step-by-step explanation:

La cantidad ahorrada bajo interés compuesto está definida por la siguiente fórmula:

$C(t) = C_{o}\cdot (1+r)^{t}$ (1)

Donde:

$C_{o}$ - Capital inicial, medido en pesos.

$C(t)$ - Capital actual, medido en pesos.

$r$ - Tasa de interés, sin unidad.

$t$ - Anualidad, medida en años.

Asumiendo que la tasa de interés compuesto es anual y si $C(t) = 500000$ , $r = 0.14$ y $t = 35$ , entonces el capital inicial es:

$C_{o} = \frac{C(t)}{(1+r)^{t}}$

$C_{o} = \frac{500000}{(1+0.14)^{35}}$

$C_{o} = 5096.83$

Se necesita depositar un capital inicial de $ 5096.83 pesos para lograr $ 500000 pesos a un interés compuesto de 14 % anual luego de 35 años.

6 0

3 years ago

Is it true that: If f"(c) > 0, then the slope of

zheka24 [161]

Answer:

yes

Step-by-step explanation:

the FIRST derivative of a function tells us the slope of a tangent line to the curve at any point. if is positive, then the curve must be increasing. If is negative, then the curve must be decreasing.

the SECOND derivative gives us the slope of the slope function (in other words how fast the slope of the original function changes, and if it is accelerating up - positive - or if it is avengers down - negative).

so, the first derivative would be fully sufficient to get the answer of if the slope of the function at that point is positive or negative.

but because it is only a "if" condition and not a "if and only if" condition, the statement is still true.

there are enough cases, where the slope is positive, but the second derivative is not > 0 (usually = 0).

but if even the second derivative is positive, then, yes, the slope of the original function must be positive too.

3 0

2 years ago

If a cone-shaped water cup holds 23 cubic inches and has a radius of 1 inch, what is the height of the cup?

schepotkina [342]

23 inches is the answer

7 0

3 years ago

A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x)equals72 com

solmaris [256]

Answer:

Part (A)

1. Maximum revenue: $450,000

Part (B)

2. Maximum protit: $192,500

3. Production level: 2,300 television sets

4. Price: $185 per television set

Part (C)

5. Number of sets: 2,260 television sets.

6. Maximum profit: $183,800

7. Price: $187 per television set.

Explanation:

0. Write the monthly cost and price-demand equations correctly:

Cost:

$C(x)=72,000+70x$

Price-demand:

$p(x)=300-\dfrac{x}{20}$

Domain:

$0\leq x\leq 6000$

1. Part (A) Find the maximum revenue

Revenue = price × quantity

Revenue = R(x)

$R(x)=\bigg(300-\dfrac{x}{20}\bigg)\cdot x$

Simplify

$R(x)=300x-\dfrac{x^2}{20}$

A local maximum (or minimum) is reached when the first derivative, R'(x), equals 0.

$R'(x)=300-\dfrac{x}{10}$

Solve for R'(x)=0

$300-\dfrac{x}{10}=0$

$3000-x=0\\\\x=3000$

Is this a maximum or a minimum? Since the coefficient of the quadratic term of R(x) is negative, it is a parabola that opens downward, meaning that its vertex is a maximum.

Hence, the maximum revenue is obtained when the production level is 3,000 units.

And it is calculated by subsituting x = 3,000 in the equation for R(x):

R(3,000) = 300(3,000) - (3000)² / 20 = $450,000

Hence, the maximum revenue is $450,000

2. Part (B) Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set. 

i) Profit(x) = Revenue(x) - Cost(x)

Profit (x) = R(x) - C(x)

$Profit(x)=300x-\dfrac{x^2}{20}-\big(72,000+70x\big)$

$Profit(x)=230x-\dfrac{x^2}{20}-72,000\\\\\\Profit(x)=-\dfrac{x^2}{20}+230x-72,000$

ii) Find the first derivative and equal to 0 (it will be a maximum because the quadratic function is a parabola that opens downward)

Profit' (x) = -x/10 + 230

-x/10 + 230 = 0

-x + 2,300 = 0

x = 2,300

Thus, the production level that will realize the maximum profit is 2,300 units.

iii) Find the maximum profit.

You must substitute x = 2,300 into the equation for the profit:

Profit(2,300) = - (2,300)²/20 + 230(2,300) - 72,000 = 192,500

Hence, the maximum profit is $192,500

iv) Find the price the company should charge for each television set:

Use the price-demand equation:

p(x) = 300 - x/20

p(2,300) = 300 - 2,300 / 20

p(2,300) = 185

Therefore, the company should charge a price os $185 for every television set.

3. Part (C) If the government decides to tax the company $4 for each set it produces, how many sets should the company manufacture each month to maximize its profit? What is the maximum profit? What should the company charge for each set?

i) Now you must subtract the $4 tax for each television set, this is 4x from the profit equation.

The new profit equation will be:

Profit(x) = -x² / 20 + 230x - 4x - 72,000

Profit(x) = -x² / 20 + 226x - 72,000

ii) Find the first derivative and make it equal to 0:

Profit'(x) = -x/10 + 226 = 0

-x/10 + 226 = 0

-x + 2,260 = 0

x = 2,260

Then, the new maximum profit is reached when the production level is 2,260 units.

iii) Find the maximum profit by substituting x = 2,260 into the profit equation:

Profit (2,260) = -(2,260)² / 20 + 226(2,260) - 72,000

Profit (2,260) = 183,800

Hence, the maximum profit, if the government decides to tax the company $4 for each set it produces would be $183,800

iv) Find the price the company should charge for each set.

Substitute the number of units, 2,260, into the equation for the price:

p(2,260) = 300 - 2,260/20

p(2,260) = 187.

That is, the company should charge $187 per television set.

7 0

3 years ago

Need help asap, thanks!

il63 [147K]

Answer:

b option

Step-by-step explanation:

Refer to the attachment.

Hope it helps :)

6 0

2 years ago