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3 years ago

Choose the function that represents the data in the table.

Mathematics

1 answer:

Usimov [2.4K]3 years ago

3 0

Given:

It is given that the function represents the data in the table.

We need to determine the function.

Slope:

The slope can be determined using the formula,

$m=\frac{y_2-y_1}{x_2-x_1}$

Let us substitute the coordinates (1,6.5) and (4,8) in the above formula, we get;

$m=\frac{8-6.5}{4-1}$

$m=\frac{1.5}{3}$

$m=0.5$

Thus, the slope is 0.5

y - intercept:

The y - intercept is the value of y when x = 0.

Thus, from the table, when x = 0 the corresponding y value is 6.

Therefore, the y - intercept is $b=6$

Equation of the function:

The equation of the function can be determined using the formula,

$y=mx+b$

Substituting the values $m=0.5$ and $b=6$ , we get;

$y=0.5x+6$

Thus, the equation of the function is $y=0.5x+6$

Hence, Option C is the correct answer.

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Select all the correct answers. In which pairs of matrices does AB = BA?

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In order to multiply a matrix by another matrix, we multiply the rows in the first matrix by the columns in the other matrix (How this is done is shown below)

To determine the pairs of matrices that AB=BA, we will determine AB and BA for each of the options below.

For the first option

$A= \left[\begin{array}{cc}1&0&-2&1&\end{array}\right]; B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times5)+(0\times3)&(1\times0)+(0\times 2)&(-2\times5)+(1\times3)&(-2\times0)+(1\times2)&\end{array}\right]\\AB= \left[\begin{array}{cc}5+0&0+0&-10+3&0+2&\end{array}\right]\\AB = \left[\begin{array}{cc}5&0&-7&2&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(5\times1)+(0\times-2)&(5\times0)+(0\times 1)&(3\times1)+(2\times-2)&(3\times0)+(1\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}5+0&0+0&3+-4&0+2&\end{array}\right]\\BA = \left[\begin{array}{cc}5&0&-1&2&\end{array}\right] \\$

∴ AB≠BA

For the second option

$A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]; B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times3)+(0\times6)&(1\times0)+(0\times -3)&(-1\times3)+(2\times6)&(-1\times0)+(2\times-3)&\end{array}\right]\\AB= \left[\begin{array}{cc}3+0&0+0&-3+12&0+-6&\end{array}\right]\\AB = \left[\begin{array}{cc}3&0&9&-6&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(3\times1)+(0\times-1)&(3\times0)+(0\times 2)&(6\times1)+(-3\times-1)&(6\times0)+(-3\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}3+0&0+0&6+3&0+-6&\end{array}\right]\\BA = \left[\begin{array}{cc}3&0&9&-6&\end{array}\right] \\$

Here AB = BA

For the third option

$A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]; B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times5)+(0\times3)&(1\times0)+(0\times 2)&(-1\times5)+(2\times3)&(-1\times0)+(2\times2)&\end{array}\right]\\AB= \left[\begin{array}{cc}5+0&0+0&-5+6&0+4&\end{array}\right]\\AB = \left[\begin{array}{cc}5&0&1&4&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(5\times1)+(0\times-1)&(5\times0)+(0\times 2)&(3\times1)+(2\times-1)&(3\times0)+(2\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}5+0&0+0&3+-2&0+4&\end{array}\right]\\BA = \left[\begin{array}{cc}5&0&1&4&\end{array}\right] \\$

Here also, AB=BA

For the fourth option

$A= \left[\begin{array}{cc}1&0&-2&1&\end{array}\right]; B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times3)+(0\times6)&(1\times0)+(0\times -3)&(-2\times3)+(1\times6)&(-2\times0)+(1\times-3)&\end{array}\right]\\AB= \left[\begin{array}{cc}3+0&0+0&-6+6&0+-3&\end{array}\right]\\AB = \left[\begin{array}{cc}3&0&0&-3&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(3\times1)+(0\times-2)&(3\times0)+(0\times 1)&(6\times1)+(-3\times-2)&(6\times0)+(-3\times1)&\end{array}\right]\\BA= \left[\begin{array}{cc}3+0&0+0&6+6&0+-3&\end{array}\right]\\BA = \left[\begin{array}{cc}3&0&12&-3&\end{array}\right] \\$

Here, AB≠BA

Hence, it is only in the second and third options that AB = BA

$A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right] B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\$ and $A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\$

Learn more on matrices multiplication here: brainly.com/question/12755004

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