How many halves are in 35/10?
1 answer:
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Looks like the system is
x + 4y - z = -14
5x + 6y + 3z = 4
-2x + 7y + 2z = -17
or in matrix form,
Cramer's rule says that
where is the solution for i-th variable, and
is a modified version of
with its i-th column replaced by
.
We have 4 determinants to compute. I'll show the work for det(A) using a cofactor expansion along the first row.
The modified matrices and their determinants are
Then by Cramer's rule, the solution to the system is
Answer:
See below
Step-by-step explanation:
We start by dividing the interval [0,4] into n sub-intervals of length 4/n
Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.
Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)
but
so the upper sum equals
When both
and
tend to zero and the upper sum tends to
Answer:
Step-by-step explanation:
From the given information.
R3 is a vector space over the field R, where R is the set of real numbers.
Where;
The set W = {(0, x2, x3): x2 and x3 are real numbers} is the subspace of ³
To proof:
(0,0,0) ∈ W ⇒ W ≠ ∅
Suppose u and v is an element of W;
i.e.
u,v ∈ W (which implies that) ⇒ u (0,x2,x3) and v = (0,y2,y3) are real numbers.
Then
u+v = (0,x2,x3) +(0, y2, y3)
u+v = (0,x2+y2, x2+y3) ∈ W
⇒ u+v ∈ W ----- (1)
Now, if we take any integer to be an element of the real number
i.e
∝ ∈
∝* = ∝(0,x2,x3)
∝* = (0,x2,x3) ∈ W
⇒ ∝*u ∈ W ------(2)
Thus from (1) and (2), W is a subspace of ³