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iogann1982 [59]
3 years ago
15

The points (3, 12) and (2,8) form a proporitonal relationship. What is the slope of the line that passes through

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
8 0

Answer:

y=4x

Step-by-step explanation:

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The equation y=−15x+12 is the slope-intercept form of which equation?
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I honestly don't know but I think it would come out to be 4/5

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Given mn, find the value of x.<br> 108
adoni [48]

Answer:

72 degrees

Step-by-step explanation:

Referencing off of line t being a straight line, a straight line is 180 degrees. If you subtract 108 from 180, you would get 72. Since this intersecting line is going through 2 paralel lines, the 2 groupings of angles reflect each other.

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If 100 units of a product cost $5,000 and 200 units cost $8,000, what is the incremental cost per unit of the second 100 units?
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I don’t know why the word “million” is there, but the incremental cost per unit of the secon hundred units is $30 per unit.
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Mrs. Smith had the following transactions during the week:
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3 years ago
Read 2 more answers
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.  

5 0
3 years ago
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