Find the slope of the following coordinate points. (4.25, 0) and (3.5, 3)

1 answer:

3 0

The slope is -4.

First you have to do 3 - 0 which is 3

Next do 3.5 - 4.25 which is -.75

The last step is to do 3 divided by -.75 which gets you the final answer which is -4.

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Angelina_Jolie [31]

Answer:

Step-by-step explanation:

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4 0

2 years ago

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How do you solve 1-5+1×(4×4-31)×8

Alja [10]

Hey there!

In order to solve this, remember PEMDAS. First, start with solving whatever is in parentheses. Then, move on to exponents. Then, complete the multiplication and division. Lastly, add or subtract anything remaining. Omit any steps that aren't present.

Parentheses:

1 – 5 + 1 × (4 × 4 – 31) × 8
1 – 5 + 1 × (–15) × 8

Multiplication/Division (Left to Right):

1 – 5 + (–15) × 8
1 – 5 + (–120)
1 – 5 – 120

Addition/Subtraction (Left to Right):

–4 – 120
–124

Your answer is –124.

Hope this helped you out! :-)

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3 years ago

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 $a_i=0$ .