All categories

4 years ago

I just need 5 please help show work and I’ll give brainliest

Mathematics

1 answer:

Firlakuza [10]4 years ago

8 0

The function is an Even function f(x)=f(-x).
Amswer:The even functions are Option 2,Option 3 and Option 6.

You might be interested in

A customer buys 17.01 in gas and requests one five dollar [$5] lottery ticket, two one dollar [$1] lottery tickets, and one [$3]

Alik [6]

17.01+5+2+3= 27.01 (total expense)
100-27.01= 72.99 (change)
5+2= 7 (total winnings)
72.99+7.00= 79.99 (total change).

4 0

4 years ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

6 0

3 years ago

I need someone to explain to me how to solve this, because I need to show my work.

masha68 [24]

So to find the answer you have to know what the mixture is, what the trend is,

on day one he used 4 cans of blue and 6 cans of yellow

day two, 6 blue and 9 yellow

both are divisible by 2 and 3 respectively, additionally when you divide the number of blue cans Marc used by the number of yellow cans each day you get a sum of .66

those are the trends

so therefore day three would have the same trend.

Now on day three Marc will have 4 cans of blue paint and 5 cans of yellow paint remaining, what I did here to find the answer was divide numbers under 4 and 5 until I got a division problem that equaled .66

my answer was 2 blue cans and 3 yellow.

The answer is that on day three the highest number of cans that Marc can mix to get his favorite shade of green is 2 blue and 3 yellow