Solve the following system using elimination:
{-2 x + 2 y + 3 z = 0 | (equation 1)
{-2 x - y + z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)
Subtract equation 1 from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x - 3 y - 2 z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)
Multiply equation 2 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)
Add equation 1 to equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{0 x+5 y + 6 z = 5 | (equation 3)
Swap equation 2 with equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+3 y + 2 z = 3 | (equation 3)
Subtract 3/5 × (equation 2) from equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - (8 z)/5 = 0 | (equation 3)
Multiply equation 3 by 5/8:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - z = 0 | (equation 3)
Multiply equation 3 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Subtract 6 × (equation 3) from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y+0 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Divide equation 2 by 5:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Subtract 2 × (equation 2) from equation 1:
{-(2 x) + 0 y+3 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
v0 x+0 y+z = 0 | (equation 3)
Subtract 3 × (equation 3) from equation 1:
{-(2 x)+0 y+0 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Divide equation 1 by -2:
{x+0 y+0 z = 1 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Collect results:
Answer: {x = 1, y = 1, z = 0
Answer: False
Step-by-step explanation:
When comparing fractions such as 2/3 and 4/5, you could also convert the fractions (if necessary) so they have the same denominator and then compare which numerator is larger.
Answer:
I thinK it's D
it might be wrong but I think it's the right answer
Answer:
6 bench seats creates an arc of 56.52 foot.
Step-by-step explanation:
We are given the following in the question:
Total number of bench seats in Ferris wheel = 24
Radius of Ferris wheel = 36 feet
We have to find the arc length created by 6 bench seats.
Angle formed by 6 bench seats =
![\theta = \dfrac{6}{24}\times 360 = 90^\circ](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cdfrac%7B6%7D%7B24%7D%5Ctimes%20360%20%3D%2090%5E%5Ccirc)
Arc length =
![\dfrac{\theta}{360}\times 2\pi r](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctheta%7D%7B360%7D%5Ctimes%202%5Cpi%20r)
Putting values, we get,
![\dfrac{90}{360}\times 2(3.14)(36) = 56.52\text{ feet}](https://tex.z-dn.net/?f=%5Cdfrac%7B90%7D%7B360%7D%5Ctimes%202%283.14%29%2836%29%20%3D%2056.52%5Ctext%7B%20feet%7D)
Thus, 6 bench seats creates an arc of 56.52 foot.
Answer:
D.) Median for both bakeries because the data is not symmetric
Step-by-step explanation:
You can compare box-and-whisker plots for the two bakeries
.
Neither bakery has a normal distribution.
The median for Bakery A is not in the middle of the box.
Rather, it cuts the box into two unequal parts, so the data does not have a normal distribution.
The longer part is below the median, so the data are skewed to the bottom.
Similarly, the data for Bakery B are skewed to the top.
The values in the tails pull the mean down for Bakery A and up for Bakery B, so it is better to describe the centres of distribution in terms of the medians.