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3 years ago

Can somebody pls help me understand this question? (Give an explanation)

Mathematics

1 answer:

emmainna [20.7K]3 years ago

3 0

Answer:

l: 1

k: 2

Step-by-step explanation:

The slope of a line is "the amount it goes up divided by the amount it goes right".

So if you pick a point on the line, and then step one grid position to the right, how much does it go up (or down, then you have a negative slope)?

In line l, it goes up one grid position for every step to the right, hence the slope is 1.

In line k, it goes up two positions for every step to the right, hence the slope there is 2.

Hope this explains it.

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What is an equation of the line that passes through the point (3,-1) and is parallel to the line 5x+3y=6

777dan777 [17]

Answer:

$y=-\frac{5}{3} x+4$

Step-by-step explanation:

Since the line we are trying to study is parallel to the one given by the standard equation 5x + 3y = 6, it needs to have the same slope as this given line has. Let's write then this equation in slope-intercept so we can find what the slope is :

$5x+3y=6\\3y=-5x+6\\y=-\frac{5}{3} x+\frac{6}{3} \\y=-\frac{5}{3} x+2$

Then the slope of our line must also be "-5/3" in order to be parallel to the given line.

Now, since we also know a point (3, -1) through which the new line should go, we use the point-slope form of a line:

$y-y_0=m(x-x_0)\\y-(-1)=-\frac{5}{3} (x-3)\\y+1=-\frac{5}{3} x+5\\y=-\frac{5}{3} x+4$

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3 years ago

12/15 and 24/30 equivalent fractions

Anton [14]

Answer:

12/15 and 24/30 are equivalent fractions

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3 years ago

simplify the expression below. Express the simplified expression in standard form (12s^4-6s^2+4s)+(6s^4-4s+27)-(4s^4+s^2+12)

joja [24]

<h3>Answer: 14s^4 - 7s^2 + 15</h3>

====================================================

Work Shown:

(12s^4-6s^2+4s)+(6s^4-4s+27)-(4s^4+s^2+12)

12s^4-6s^2+4s+6s^4-4s+27-4s^4-s^2-12

(12s^4+6s^4-4s^4)+(-6s^2-s^2)+(4s-4s)+(27-12)

14s^4-7s^2+0s+15

14s^4-7s^2+15

Note: don't forget to distribute the negative to every term in the last parenthesis

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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 $a_i=0$ .

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