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3 years ago

HELP ASAP PLEASE! GIVING AWAY 30 POINTS AND BRAINLIEST!

Mathematics

2 answers:

Papessa [141]3 years ago

7 0

Answer:

DB!

Step-by-step explanation:

Question a: 3+3+2+2 = 10 dinners

Question b: 15-30 and 30-45, because most people choose that range.

Hope I helped!

mezya [45]3 years ago

7 0

Answer:

a) 10 b) 2-3 dollars

Step-by-step explanation:

a) 3+3+2+2

b) (2+2+3+3)/2 equal 2.5 mean, range between them is 2 to 3

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Write 7*7*b*b*b in exponential form

ycow [4]

Answer:

7²b³

Step-by-step explanation:

An exponent signifies repeated multiplication. Here, the factor 7 is repeated twice, and the factor b is repeated 3 times. These will get exponents of 2 and 3, respectively:

7²b³

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3 years ago

Find the value of x then classify the triangle

Sliva [168]

Answer: 88º

Step-by-step explanation:

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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 $a_i=0$ .