I just took the test, the answer is: Fiona's meals are less expensive. :)
Answer:
0.0001 kilometers I hope I helped
Answer:

Step-by-step explanation:
we have to orthonormalize the vectors:

According to Gram - Schmidt process, we have:
where, 
The normalized vector is: 
Now, the first step.
= u₁
Therefore, e₁ = 

Now, we find e₂.


Therefore, 

To find e₃:




So, we have the orthonormalized vectors
.
Hence, the answer.
Answer:
Step-by-step explanation:
f(x) = 2x + 11
f(-3) = 2(-3) + 11
= - 6 + 11 = 5
Answer:
4y^2 x (6x^3 - 4xy - 3y)
Step-by-step explanation:
Factor out -4y^2 from the expression