Well, take a look at this
-10x + 3 = -x + 21
+ 10 x +10x
--------------------------
3 = 9x + 21
-21 -21
---------------------
-18 = 9x
-2 = x
hope this helps
Step-by-step explanation:
for the table on the left (Q5)
y(-1) = -5
y(1) = -1
y(3) = 3
for the table on the left (Q7)
y(0) = 5
y(1) = 2
y(2) = -1
y(3) = -4
Answer:
it's 0
Step-by-step explanation:
probably..............
(46.6 -47)/4 = -1
(47.4-47)/4 = 1
So, +/- 1 std, that is 95% So, (D)!
Answer:
S is not the subspace of 
Step-by-step explanation:
Let us suppose two vectors u and v belong to the S such that the property of xy≥0 is verified than
![v=\left[\begin{array}{c}0 \\1\end{array}\right]](https://tex.z-dn.net/?f=v%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%20%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
Both the vectors satisfy the given condition as follows and belong to the S

Now S will be termed as subspace of R2 if
- u+v also satisfy the condition
- ku also satisfy the condition
Taking u+v
![u+v=\left[\begin{array}{c}-1 \\0\end{array}\right]+\left[\begin{array}{c}0 \\1\end{array}\right]\\u+v=\left[\begin{array}{c}-1+0 \\0+1\end{array}\right]\\u+v=\left[\begin{array}{c}-1 \\1\end{array}\right]](https://tex.z-dn.net/?f=u%2Bv%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%20%5C%5C0%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%20%5C%5C1%5Cend%7Barray%7D%5Cright%5D%5C%5Cu%2Bv%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%2B0%20%5C%5C0%2B1%5Cend%7Barray%7D%5Cright%5D%5C%5Cu%2Bv%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%20%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
Now the condition is tested as

This indicates that the condition is not satisfied so S is not the subspace of 