PLEASE HELP

1 answer:

4 0

I believe this is the right way to write the function

N = [log A − log P] / log(1.0175)

N =[ log (1210.26) - log (1000)]/log (1.0175)

N =log [1210.26/1000] / log (1.0175) = log (1.21026)/log(1.0175) = 0.08288/0.00753 = 11

N = 11 months

You might be interested in

Three vertices of a rectangle are (–2, 1), (–2, –3), and (4, –3)

Tpy6a [65]

The answer is C
it's easier if you graph it
we know that the x of the unknown vertice is 4, because this is a rectangle and that vertices has to be on the same x line as the point below.
y of the unknown vertice has to be 1 because the vertice to the left needs to be on the same y line as the unknown vertice

6 0

3 years ago

Hawfields won 17 out of 20 basketball games. What percent of the games did they win?

timama [110]

The hawfields won 85% of the games :)

8 0

3 years ago

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1

AysviL [449]

Answer:

a) The system has a unique solution for $k\neq 6$ and any value of $h$ , and we say the system is consisted

b) The system has infinite solutions for $k=6$ and $h=8$

c) The system has no solution for $k=6$ and $h\neq 8$

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms ( $h$ ), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

$\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right]$

The we can use the transformation $r_0\rightarrow r_0 -2r_1$ , obtaining:

$\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right]$ .

Now we can start the analysis:

If $k\neq 6$ then, the system has a unique solution for any value of $k$ , meaning that the last row will transform back to the equation as:

$(k-6)x_2=h-8\\x_2=h-8/(k-6)$

from where we can see that only in the case of $k=6$ the value of $x_2$ can not be determined.

if $k=6$ and $h=8$ the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=0$ which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get $x_1$ as a function of $x_2$ o viceversa. Thus, $x_2$ ( $x_1$ ) is called a parameter since there are no constraints on what values they can take on.

if $k=6$ and $h\neq 8$ the system has no solution. Again by substituting in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=h-8$ which is false for all values of $h\neq 8$ and since we have something that is not possible $(0\neq h-8,\ \forall \ h\neq 8)$ the system has no solution