The table doesn't represent linear function
Step-by-step explanation:
We need to identify if the table represent a linear function or not.
<u>Linear Function </u>
A linear function is defined as a straight line with an x and y intercept and the same slope through the whole line.
Finding the slope of elements in the table:
x y
0 0
1 1
2 8
3 27
Slope= y/x
Slope = 0/0=0
slope = 1/1 = 1
slope = 8/2 = 2
slope = 27/3 = 9
The function represented is: y=x^3
Since the slope of points x and y in the table is not same, and its graph is not linear.
So, the table doesn't represent linear function
Keywords: Linear Function
Learn more about Linear function at:
#learnwithBrainly
Answer:
cosjk = √55 i/3
tanjk = 8/√55 i
Step-by-step explanation:
Given
sin jk = 8/3
According to SOH CAH TOA
Sin theta = opposite/hypotenuse = 8/3
Opposite = 8
hypotenuse = 3
Get the adjacent using the pythagoras theorem
hyp² = opp²+adj²
adj² = hyp² - opp²
adj² = 3² - 8²
adj² = 9-64
adj² = -55
adj = √-55
adj = √55 i (i = √-1)
Get cosjk
cosjk = adj/hyp
cosjk = √55 i/3
Get tanjk
tanjk = opp/adj
tanjk = 8/√55 i
Answer:
It can be filled 15 times
Step-by-step explanation:
If 5 casks hold 120 gallons each,
5x120= 600
To find how many times a 40gallon barrel can be filled,
= 600/40
= 15
This means a 40 gallon barrel will be filled 15 times.
Answer:
x= -3 and y= 0
Step-by-step explanation:
5x+2y=-15
<u>2x-2y=-6 </u>
<u>7x =-21</u>
x= -3
Putting value of x in equation 1
5(-3) +2y=-15
-15+2y= -15
2y= 0
y= 0
This can be solved with the help of matrices
In matrix form the above equations can be written in the form
![\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%262%5C%5C2%26-2%5C%2F%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x\\y\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-15%5C%5C-6%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Let
= A = X and = B
Then AX= B
or X= A⁻¹ B
where A⁻¹= adj A/ ║A║ where mod A≠ 0
adj A= ![\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26-2%5C%5C-2%265%5C%2F%5Cend%7Barray%7D%5Cright%5D)
║A║= ( 5*-2- 2*2)= -10-4= -14≠0
X= A⁻¹ B
=- 1/14
=- 1/14 ![\left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%2A-15%26%2B%20-2%2A-6%5C%5C-2%2A-15%26%2B%205%2A-6%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=- 1/14 ![\left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%2030%26%2B12%5C%5C30%26%2B-30%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=- 1/14 ![\left[\begin{array}{ccc}42\\0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D42%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-42%2F14%5C%5C0%2F-14%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-3\\0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
From here x= -3 and y= 0
Solution Set = [(-3,0)]
