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galben [10]
3 years ago
9

Solve for x: 1 over 3 (2x − 8) = 4.

Mathematics
2 answers:
Elena-2011 [213]3 years ago
7 0
The answer is c because since twenty minus eight is twelve one third times twelve is four
S_A_V [24]3 years ago
3 0
The correct answer is:  [C]:  "10" . 
____________________________________________________
    →  " x = 10 "  .
____________________________________________________
Explanation:
____________________________________________________
 Given:
____________________________________________________
    →   (1/3) * (2x − 8) = 4  ;  Solve for "x" ; 
____________________________________________________
To get rid of the "fraction" ; multiple BOTH SIDE of the equation by "3" : 

   →   3 * { (1/3) * (2x − 8) } = 3 * 4 ; 

to get: 

 1(2x − 8) = 12 ; 

↔  2x − 8 = 12 ; 

Add "8" to BOTH SIDES of the equation ; 

 →  2x − 8 + 8 = 12 + 8 ;

to get:

→  2x = 20 ; 

Divide EACH SIDE of the equation by "2" ; 
 to isolate "x" on one side of the equation; & to solve for "x" 

→ 2x / 2  = 20/ 2 ; 

→  x = 10  ;  which is:  Answer choice:  [C]:  "10" .
________________________________________________
Let us check our answer:  " x = 10" ;  by plugging in "10" for all values of "x" into the original equation; to see if the answer holds true for the equation ; that is, to see if both sides of the equation are equal when "x = 10" ; 

The original equation:  " (1/3) * (2x − 8) = 4 " .

    →  (1/3) * [2(10) − 8)  =?  4 ?? ;

    →  (1/3) * [20 − 8)  =?  4 ?? ; 

    →  (1/3) * (12)  =?  4 ?? ;   Yes!
_____________________________________________
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All vectors are in Rn. Check the true statements below:
Oduvanchick [21]

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

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 ||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}. Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then ||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||.

C) Consider S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2. This set is orthogonal because (0,2)\cdot(2,0)=0(2)+2(0)=0, but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in \mathbb{R}^n. Then the columns of A form an orthonormal set. We have that A^{-1}=A^t. To see this, note than the component b_{ij} of the product A^t A is the dot product of the i-th row of A^t and the jth row of A. But the i-th row of A^t is equal to the i-th column of A. 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 A^t A=I    

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 \{u_1,u_2\cdots u_p\} and suppose that there are coefficients a_i such that a_1u_1+a_2u_2\cdots a_nu_n=0. 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 a_i||u_i||=0 then a_i=0.  

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