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3 years ago

Need help on this, I have no idea what should go on the first line or the second line

Mathematics

1 answer:

klasskru [66]3 years ago

3 0

Answer:

1.) 120

2.)120

Step-by-step explanation:

1.) Do length times width

2.) For the number 120 you don't have to round, so it just stays the same

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Mr.diaz’s class is taking a field trip to the science museum there are 23 students in the class and a student admission ticket i

PIT_PIT [208]

since there are 23 students and a ticket costs $8 for each you would:

23x8=184

the student's tickets cost $184 all together

hope this helps! :3

4 0

4 years ago

What is the diameter of this circle if the circumference is 30 pi inches?

djyliett [7]

Answer:

30 inches

Step-by-step explanation:

Diameter is equal to two times the radius. The circumference formula is 2piR, and you can cancel the pi's. The answer would already be in diameter form.

8 0

3 years ago

Read 2 more answers

Which algebraic expression means “four less than three times a number”?

Monica [59]

It’s 3n-4

I might be right or wrong

5 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 $a_i=0$ .