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Annette [7]
3 years ago
11

If the day tomorrow is sunday, what was it yesterday?

Mathematics
2 answers:
Basile [38]3 years ago
8 0
Tommorow is sunday, meaning it is currently saturday. So yesterday was friday.
WARRIOR [948]3 years ago
6 0
Friday isn't it? cos if today was Saturday yesterday would be Friday and tomorrow would be sunday
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Can u guys help pls?
Contact [7]

Answer:

(2 {n}^{2} ) ^{4}  \\ 2 \times 2 \times 2 \times 2 \times  {n}^{2 \times 4}  \\  {2}^{4}  \times  {n}^{8}  \\  = 16 {n}^{8}

<h3>Answer B is correct</h3>

7 0
3 years ago
2 ( 4 x + 4 ) + 6 ( 2 x + 6 ) + 4 = 20 x + 48 solving for x
Alisiya [41]

Answer:

infinitely many solutions

Step-by-step explanation:

8x+8+12x+36+4=20x+48

20x + 48 = 20x + 48

0= 0

3 0
3 years ago
5 (10k + 1) + 2 (2 + 8k)
Eddi Din [679]

Answer:

66k+9

Step-by-step explanation:

Simplify: 50k+5+4+16k

Combine like-terms: 66k+9

Hope this helps

6 0
3 years ago
Read 2 more answers
X+3=2X+6(x+4)<br> Please Answer!
nataly862011 [7]
X=10
-3 on both sides
Then -6(x+4)+3 to the other side
4 0
2 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
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