Classify this number as rational, irrational, or nether: 6

aalyn [17]

Answer:

(-3,0)

Step-by-step explanation:

the vertex is where the quadratic function reaches its maximum or minimum height and as you can see on this graph the maximum height of this equation is at the (-3,0) point meaning that (-3,0) is the vertex

5 0

2 years ago

The accompanying table shows the number of bacteria present in a certain culture over a 4 hour period, where x is the time, in h

Vlad [161]

Answer:

Step-by-step explanation:

I really hope you know how to use your calculator to find regression equations, because in this forum it is impossible to teach you how to do that. I ran the data through my calculator and got the exponential equation as

$y=1963.56(1.14)^x$ which means at 4 hours, we started with 1963 bacteria and it is growing at a rate of 114% each hour. Since we started out with an x value of 0, and that value represents the number of bacteria after 4 hours, then time 0 is 4 hours, which means that time 12 is 16 hours. We replace the x in the equation with a 12 and do the math:

$y=1963.56(1.14)^{12$ which gives us, in the end,

y = 9460

4 0

2 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