Answer: 1/4x + 1/2x = 12
x = 16
Step-by-step explanation: combine the 1/4x and 1/2x to 3/4x because they are like terms then divide 12 by 3/4 or 0.75 to get 16
Answer: 160 miles
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Explanation:
x = number of miles driven, y = cost
First plan: y = 0.08x + 61.98
Second plan: y = 0.13x + 53.98
Equate the two right hand sides of each equation; solve for x
0.13x + 53.98 = 0.08x + 61.98
0.13x - 0.08x = 61.98 - 53.98
0.05x = 8
x = 8/0.05
x = 160
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Extra Info: plugging x = 160 into each equation gives us...
y = 0.08x + 61.98 = 0.08*160+61.98 = 74.78
y = 0.13x + 53.98 = 0.13*160 + 53.98 = 74.78
Therefore, driving 160 miles for each plan yields the same cost $74.78, which helps us confirm we have the right answer.
A really easy way to do this would be converting them all to decimals by dividing the top number by the bottom number.
7/5 = 1.4
15/4 = 3.75
3/2 = 1.5
11/4 = 2.75
13/3 = 4.33
Start with the whole numbers. 3/2 and 7/5 must go first since they both have a 1. 7/5 goes before 3/2 since it has a 4 in the tenths place, while 3/2 has a 5.
Next would be 11/4 since it has a 2, then 15/4 with a 3, and lastly 13/3 with a 4 as a whole number
Answer is 7/5 , 3/2 , 11/4, 15/4 , 13/3
Answer:
A), B) and D) are true
Step-by-step explanation:
A) We can prove it as follows:
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 
.
