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2 years ago

You plan to accumulate 100,000 at the end of 42 years by making the follow-

Mathematics

1 answer:

Mariana [72]2 years ago

7 0

Answer:

Y = 479.17

Step-by-step explanation:

At the end of year 14, the balance from the deposits of X can be found using the annuity due formula:

A = P(1+r/n)((1 +r/n)^(nt) -1)/(r/n)

where P is the periodic payment, n is the number of payments and compoundings per year, t is the number of years, and r is the annual interest rate.

A = X(1.07)(1.07^14 -1)/0.07 ≈ 24.129022X

This accumulated amount continues to earn interest for the next 28 years, so will further be multiplied by 1.07^28. Then the final balance due to deposits of X will be ...

Ax = (24.129022X)(1.07^28) = 160.429967X

The same annuity due formula can be used for the deposits of Y for the last 10 years of the interval:

Ay = Y(1.07)(1.07^10 -1)/.07 = 14.783599Y

Now we can write the two equations in the two unknowns:

Ax +Ay = 100,000

X - Y = 100

From the latter, we have ...

X = Y +100

So the first equation becomes ...

160.429967(Y +100) +14.783599Y = 100000

175.213566Y +16,043.00 = 100,000

Y = (100,000 -16,043)/175.213566 ≈ 479.17

Y is 479.17

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Write an equation for a quadratic function that has x intercepts (-3, 0) and (5, 0)

Blizzard [7]

Answer:

One possible equation is $f(x) = (x + 3)\, (x - 5)$ , which is equivalent to $f(x) = x^{2} - 2\, x - 15$ .

Step-by-step explanation:

The factor theorem states that if $x = x_{0}$ (where $x_{0}$ is a constant) is a root of a function, $(x - x_{0})$ would be a factor of that function.

The question states that $(-3,\, 0)$ and $(5,\, 0)$ are $x$ -intercepts of this function. In other words, $x = -3$ and $x = 5$ would both set the value of this quadratic function to $0$ . Thus, $x = -3\!$ and $x = 5\!$ would be two roots of this function.

By the factor theorem, $(x - (-3))$ and $(x - 5)$ would be two factors of this function.

Because the function in this question is quadratic, $(x - (-3))$ and $(x - 5)$ would be the only two factors of this function. In other words, for some constant $a$ ( $a \ne 0$ ):

$f(x) = a\, (x - (-3))\, (x - 5)$ .

Simplify to obtain:

$f(x) = a\, (x + 3)\, (x - 5)$ .

Expand this expression to obtain:

$f(x) = a\, (x^{2} - 2\, x - 15)$ .

(Quadratic functions are polynomials of degree two. If this function has any factor other than $(x - (-3))$ and $(x - 5)$ , expanding the expression would give a polynomial of degree at least three- not quadratic.)

Every non-zero value of $a$ corresponds to a distinct quadratic function with $x$ -intercepts $(-3,\, 0)$ and $(5,\, 0)$ . For example, with $a = 1$ :