Ask question

Ask question

All categories

3 years ago

11

I cant figure out the answer.

2" alt="1472 divide by 32" align="absmiddle" class="latex-formula">

2 answers:

Anuta_ua [19.1K]3 years ago

7 0

Answer:

<h2>46</h2><h2 /><h2 /><h2>Please leave a rating if this helped! Have a great day!</h2>

melomori [17]3 years ago

3 0

<h3>1472÷32</h3><h3> and the answer is 46 </h3>

please mark this answer as brainlist

You might be interested in

Have do you solve -4-1

ad-work [718]

Answer:

-5

Step-by-step explanation:

add them together (-4)+(-1) make a t chart it'll help you out

7 0

3 years ago

Read 2 more answers

Given: Q = 7m + 3n, R = 11 - 2m, S = n + 5, and T = -m - 3n + 8.

Vesna [10]

Q = 7m + 3n
R = 11 - 2m
S = n + 5
T = -m - 3n + 8

simplify Q + S - T

7m + 3n + n + 5 - (-m - 3n + 8) =
7m + 4n + 5 + m + 3n - 8
8m + 7n - 3 <====

7 0

3 years ago

Read 2 more answers

Evaluate (13+2) Divided by(9-4)

skelet666 [1.2K]

Answer:

=3

Step-by-step explanation:

following the order of operations, whatever is in the parentheses gets done first. so you would add 13 + 2 and get 15. then you would subtract 9-4 and get 5. then you would divide the numbers that you got and get 3.

6 0

4 years ago

Read 2 more answers

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

4 years ago

Donna's recipe for chocolate tuffles requires a ratio of 60 cups of flour to 12 cups of salt to 36 cups of sugar. how many total

Sedbober [7]

Answer:

4.8 Cups of salt and 14.4 cups of sugar

Step-by-step explanation:

5 0

4 years ago