200,195,190,185 arithmetic or geometric

algol13

When you see questions of this nature, test the individual inequalities and look out for their intersection.

For
$y < \frac{2}{3} x$
Choose a point in the lower or upper half plane created by the line
$y = \frac{2}{3} x$
The above line is the one which goes through the origin.

Now testing (1,0) yields,

$0 < \frac{2}{3} (1)$
That is,

$0 < \frac{2}{3}$
This statement is true. So we shade the lower half of
$y = \frac{2}{3} x$

For
$y \geqslant - x + 2$
We test for the origin because, it is not passing through the origin.

$0 \geqslant - (0) + 2$
This yields
$0 \geqslant 2$
This statement is false so we shade the upper half.

The intersection is the region shaded in B. The top right graph

8 0

3 years ago

Help please!!!

notka56 [123]

X=0. Y=-2. There is only one x value for this question.

5 0

3 years ago

For numbers 19 and 20 you have to solve for the missing angle.

Illusion [34]

Answer:19. 17 degrees

20. 49 degrees

Step-by-step explanation:

Because of the postulate 50+?=67

So ?=17

And 20.

94+?=143

So

?=49

(Edit)I though you only meant 19 and 20

21. X=8

22. X=12

23.x=8

24.x=7

7 0

3 years ago

a pay for service internet company charges 5 per hour for the first 3 hours of Service Plus a $10 connection fee

Marysya12 [62]

If you are charging 5 per hour use h to determine how many hours and add 10 for the connection fee

5h+10

3 0

3 years ago

Find the inverse of the given matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

BabaBlast [244]

Answer:

$A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]$

Step-by-step explanation:

We want to find the inverse of $A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Subtract row 1 multiplied by 4 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Add row 1 multiplied by 4 to row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]$

Subtract row 2 multiplied by 2 from row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]$

Divide row 3 by −27

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Add row 3 multiplied by 2 to row 1

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Subtract row 3 multiplied by 11 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0

3 years ago