6.

1 answer:

3 0

To solve this we are going to use the formula for future value: $FV=PV(1+ \frac{r}{n} )^{nt}$
where
$FV$ is the future value
$PV$ is the present value
$r$ is the interest rate in decimal form
$n$ is the number of times the interest is compounded per year
$t$ is the time in years

We know for our problem that $FV=12300$ , $r= \frac{4}{100} =0.04$ , and $t=4$ . Since the interest is compounded quarterly, it is compounded 4 times per year; therefore, $n=4$ . Lets replace those values in our formula to find and solve for $PV$ :
$FV=PV(1+ \frac{r}{n} )^{nt}$
$12300=PV(1+ \frac{0.04}{4} )^{(4)(4)}$
$PV= \frac{12300}{(1+ \frac{0.04}{4} )^{(4)(4)} }$
$PV=10489.70$

We can conclude that the present amount needed to have $12,300 after 4 years according to your given choices is <span>$10,489.69</span>

You might be interested in

(x^2-3x+2)^2 - 4x + 2

Arturiano [62]

The answer is (x-2)^2(x-1)^2-4x+2

8 0

3 years ago

A line passes through the point (6,-5) and has a slope of -5/2.

Leto [7]

Y2-Y1 over X2-X1 which gives you 7/-11

5 0

3 years ago

Question

krok68 [10]

Answer:

Acute angle between the two planes: approximately $43^\circ$ .

Step-by-step explanation:

Find the normal vector of each plane:

The normal vector of the plane $x - 2\, y + 5\, z = 3$ is $\displaystyle \begin{bmatrix}1 \\ -2 \\ 5\end{bmatrix}$ .
The normal vector of the plane $2\, x + y - 3\, z = 15$ is $\displaystyle \begin{bmatrix}2 \\ 1\\ -3\end{bmatrix}$ .

As the name suggests, there is a $90^\circ$ angle between a plane and its normal vector. The following four angles will correspond to the vertices of a quadrilateral:

The $90^\circ$ angle between the first plane and its normal vector.
The angle between the normal vector of each plane.
The $90^\circ$ angle between the second plane and its normal vector.
The smallest angle between these two planes.

The sum of these four angles should be $360^\circ$ . Two of these four angles were known to be $90^\circ$ . Once the third angle (the angle between the two normal vectors) is found, subtractions would give the measure of the other angle (the smallest angle between these two planes.)

Make use of the dot product to find the angle between these two normal vectors. Let $\theta$ denote the angle between these two vectors.

$\displaystyle \cos \theta = \frac{\begin{bmatrix}1 \\ -2 \\ 5\end{bmatrix} \cdot \begin{bmatrix}2 \\ 1 \\ -3\end{bmatrix}}{\sqrt{1^2 + (-2)^2 + 5^2} \cdot \sqrt{2^2 + 1^2 + (-3)^2}} \approx -0.73193$ .

Before continuing, notice that the smallest angle between the two planes would be $360^\circ - 90^\circ - 90^\circ - \theta = 180^\circ - \theta$ .

Consider the identity: $\cos\left(180^\circ - \theta\right) = -\cos \theta$ .

In other words, $\cos\left(180^\circ - \theta\right)$ , the cosine of the smallest angle between the two planes (which the question is asking for) will be the opposite of $\cos \theta$ , the cosine of the angle between the two normal vectors.