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3 years ago

What's the answer...I'm kinda lost

Mathematics

2 answers:

Lorico [155]3 years ago

4 0

A reflection over the x axis means y valuable will be the opposite. C

kow [346]3 years ago

3 0

Answer:

it would be D

Step-by-step explanation:

because you are reflecting the shape across the x axis

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Wai Kit had $45 more than Cheryl and 0.25 of what Zheng Yi had. If they had $510 altogether, how much less money did Cheryl have

GenaCL600 [577]

Step-by-step explanation:

Wai 1unit+45

Cheryl 1unit

Zheng 4units+180

Altogether 6units+225=510

6units = 510-225

6units=285

1unit=285÷6

= $47.5

So Wai =47.5+45 = 92.5

Cheryl =47.5

Zheng =4×47.5+180 =370

370-47.5=$322.5

Answer $322.5

4 0

2 years ago

Which choices are equivalent to the expression below?<br> Check all that apply.

gavmur [86]

There’s no answer choices..

5 0

3 years ago

9.(06.04 MC)

alekssr [168]

Answer:

took the test and its 0.5

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Do u just plot the numbers on the graph like (1,26)???

kvv77 [185]

Answer:

Yes

Step-by-step explanation:

Make sure you plot each number on the right axis. :)

4 0

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 $a_i=0$ .

5 0

3 years ago