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Masja [62]
3 years ago
9

Okay so I’m doing graph questions, so what’s “g(x) = x + 1” and what are the blanks?

Mathematics
2 answers:
nevsk [136]3 years ago
8 0

x   |   y

-2  |  -1

-1   |  0

0   |   1

1    |   2

2   |   3

do you like anime

alukav5142 [94]3 years ago
3 0

Answer:

g(0) = 1, g(1) = 2, g(2) = 3, g(3) = 4, g(4) = 5

Step-by-step explanation:

g(x) is a function meant to modify the variable x.

Basically you take the x value in the table that you want to modify and you plug it into the g(x) function.

If you want 0, you do g(0) = (0) + 1 = 1.

Or if you want 3, you do g(3) = (3) + 1 = 4

Hope this helped!

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Jada paints a circular table that has a diameter of 44 inches. What is the area of the table? Round to the nearest whole number!
Elza [17]

Answer: 1521

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5 0
3 years ago
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Helena wrote the equation using point-slope form for the line that passes through the points (5, 1) and (3, 5). Analyze the step
viva [34]

Answer:

In step 2, she didn’t use an x and y from the same coordinate pair

Step-by-step explanation:

see the attached figure to better understand the problem

we have the points (5, 1) and (3, 5)

step 1

Find the slope

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

substitute

m=\frac{5-1}{3-5}

m=\frac{4}{-2}

m=-2

step 2

The equation of the line in point slope form is

y-y1=m(x-x1)

take the point (3,5)

m=-2

substitute

y-5=-2(x-3)

y-5=-2x+6

y=-2x+6+5

step 3

y=-2x+11

therefore

In step 2, she didn’t use an x and y from the same coordinate pair

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3 years ago
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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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DIA [1.3K]

Answer:

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