Answer:
Ruth is x+4=7/3+4, esmerelda is 5x+4=35/3+4
Step-by-step explanation:
if Ruth is x, esmerelda is 5x,
four years ago,
Ruth is x+4, esmerelda is 5x+4, depend on the sum, we get:
x+4+5x+4=22, x=7/3
so:
Ruth is x+4=7/3+4, esmerelda is 5x+4=35/3+4
Answer:
A), B) and D) are true
Step-by-step explanation:
A) We can prove it as follows:
![Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y](https://tex.z-dn.net/?f=Proy_%7Bcv%7Dy%3D%5Cfrac%7B%28y%5Ccdot%20cv%29%7D%7B%7C%7Ccv%7C%7C%5E2%7Dcv%3D%5Cfrac%7Bc%28y%5Ccdot%20v%29%7D%7Bc%5E2%7C%7Cv%7C%7C%5E2%7Dcv%3D%5Cfrac%7B%28y%5Ccdot%20v%29%7D%7B%7C%7Cv%7C%7C%5E2%7Dv%3DProy_%7Bv%7Dy%20)
B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that
. Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then
.
C) Consider
. This set is orthogonal because
, but S is not orthonormal because the norm of (0,2) is 2≠1.
D) Let A be an orthogonal matrix in
. Then the columns of A form an orthonormal set. We have that
. To see this, note than the component
of the product
is the dot product of the i-th row of
and the jth row of
. But the i-th row of
is equal to the i-th column of
. If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then
E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.
In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set
and suppose that there are coefficients a_i such that
. For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then
then
.
<h2>
Answer:</h2>
We will use substitution to solve this system for x.
![x - 2y = 27\\y = 5x\\\\x - 2(5x) = 27\\\\x - 10x = 27\\\\-9x = 27\\\\x = -3](https://tex.z-dn.net/?f=x%20-%202y%20%3D%2027%5C%5Cy%20%3D%205x%5C%5C%5C%5Cx%20-%202%285x%29%20%3D%2027%5C%5C%5C%5Cx%20-%2010x%20%3D%2027%5C%5C%5C%5C-9x%20%3D%2027%5C%5C%5C%5Cx%20%3D%20-3)
Now that we know what x is, we can determine y by pluging in -3 for x in one of the original equations. I will use the first one.
![-3 - 2y = 27\\\\-2y = 30\\\\y = -15\\](https://tex.z-dn.net/?f=-3%20-%202y%20%3D%2027%5C%5C%5C%5C-2y%20%3D%2030%5C%5C%5C%5Cy%20%3D%20-15%5C%5C)
We now have our solution.
D. ![(-3, -15)](https://tex.z-dn.net/?f=%28-3%2C%20-15%29)
We can check our answer by pluging the solution into one of the original equations to see if both sides equal each other. I will use the first equation.
![-3 - 2(-15) = 27\\\\-3 + 30 = 27\\\\27 = 27](https://tex.z-dn.net/?f=-3%20-%202%28-15%29%20%3D%2027%5C%5C%5C%5C-3%20%2B%2030%20%3D%2027%5C%5C%5C%5C27%20%3D%2027)
Since the equation checks out, our solution is correct.
Answer:
Exact Form:
x= -3/2
Decimal Form:
x= -1.5
Mixed Number Form
x= -1 1/2
Step-by-step explanation:
Answer:no no you got it keep going your on the right path
Step-by-step explanation: