12 I’m pretty sure because you have x then 3 then 12
Answer:
132cm^2
Step-by-step explanation:
first break the shape up into component shapes. the shape is made of two rectangles and two triangles. the area for a rectangle is A=BH where b is base and h is height. the area of a right triangle is BH/2 where b is the base and h is the height. you are given the base and height of the top rectangle 4x7. To find the area of the triangles and the second rectangle you have to do some algebra. the total base length is 19cm and the top rectangle has a length of 7cm so 19-7 leaves you with 12cm to spare and there is two symmetric triangles on each side so 12/2 is 6. now we know all the measurements and can solve.
namely, let's rationalize the denominator in the fraction, for which case we'll be using the <u>conjugate</u> of that denominator, so we'll multiply top and bottom by its <u>conjugate</u>.
so the denominator is 5 + i, simply enough, its conjugate is just 5 - i, recall that same/same = 1, thus (5-i)/(5-i) = 1, and any expression multiplied by 1 is just itself, so we're not really changing the fraction per se.

Answer:

Step-by-step explanation:
As the given Augmented matrix is
![\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D9%26-2%260%26-4%26%3A8%5C%5C0%267%26-1%26-1%26%3A9%5C%5C8%2612%26-6%265%26%3A-2%5Cend%7Barray%7D%5Cright%5D)
Step 1 :
↔
![\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26-14%266%26-9%26%3A10%5C%5C0%267%26-1%26-1%26%3A9%5C%5C8%2612%26-6%265%26%3A-2%5Cend%7Barray%7D%5Cright%5D)
Step 2 :
↔
![\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26-14%266%26-9%26%3A10%5C%5C0%267%26-1%26-1%26%3A9%5C%5C0%26124%26-54%2677%26%3A-82%5Cend%7Barray%7D%5Cright%5D)
Step 3 :
↔
![\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26-14%266%26-9%26%3A10%5C%5C0%261%26-%5Cfrac%7B1%7D%7B7%7D%20%26-%5Cfrac%7B1%7D%7B7%7D%20%26%3A%5Cfrac%7B9%7D%7B7%7D%20%5C%5C0%26124%26-54%2677%26%3A-82%5Cend%7Barray%7D%5Cright%5D)
Step 4 :
↔
,
↔
![\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%264%26-11%26%3A-8%5C%5C0%261%26-%5Cfrac%7B1%7D%7B7%7D%20%26-%5Cfrac%7B1%7D%7B7%7D%20%26%3A%5Cfrac%7B9%7D%7B7%7D%20%5C%5C0%260%26-%20%5Cfrac%7B254%7D%7B7%7D%20%26%5Cfrac%7B663%7D%7B7%7D%20%26%3A-%5Cfrac%7B1690%7D%7B7%7D%20%5Cend%7Barray%7D%5Cright%5D)
Step 5 :
↔
![\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%264%26-11%26%3A-8%5C%5C0%261%26-%5Cfrac%7B1%7D%7B7%7D%20%26-%5Cfrac%7B1%7D%7B7%7D%20%26%3A%5Cfrac%7B9%7D%7B7%7D%20%5C%5C0%260%261%26-%5Cfrac%7B663%7D%7B254%7D%20%26%3A-%5Cfrac%7B1690%7D%7B254%7D%20%5Cend%7Barray%7D%5Cright%5D)
Step 6 :
↔
,
↔
![\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%260%26-%5Cfrac%7B71%7D%7B127%7D%20%26%3A%5Cfrac%7B176%7D%7B127%7D%20%5C%5C0%261%260%26-%5Cfrac%7B131%7D%7B254%7D%20%26%3A%5Cfrac%7B284%7D%7B127%7D%20%5C%5C0%260%261%26-%5Cfrac%7B663%7D%7B254%7D%20%26%3A%5Cfrac%7B845%7D%7B127%7D%20%5Cend%7Barray%7D%5Cright%5D)
∴ we get

Answer:
0
Step-by-step explanation:
A horizontal line has a slope of zero and a vertical line has an undefined slope