All categories

2 years ago

5th grade math. Correct answer will be marked brainliest

Mathematics

1 answer:

Ivanshal [37]2 years ago

7 0

Answer:

3,5

Step-by-step explanation:

You might be interested in

Please help due tomorrow, multiple choice

anzhelika [568]

Answer:

Step-by-step explanation:

12 is F

6 0

3 years ago

Please help me please!!

irakobra [83]

Answer:

mok

Step-by-step explanation:

mok

7 0

3 years ago

Hey, try this one out guys.

MrRissso [65]

Answer:

879cm

Step-by-step explanation:

Radius is 35cm.

Diameter is 70cm.

Pie/Pi is 3.14 and so on.

Perimeter of circle:70×3.14=219.8

219.8×4=879.2

Round Off:879.2cm

6 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

Can anyone tell me if this is right

Rashid [163]

Answer:

yes it is correct

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers