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3 years ago

Pls help with these two math questions

Mathematics

2 answers:

densk [106]3 years ago

7 0

I see it’s math but 1-1= 0

-BARSIC- [3]3 years ago

4 0

Number 18 is D) (3+8m)-7

Number 19 is:

1) z>1

2) z>-1

3) z<1

4) z<-1

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Which ordered pair is a solution of the system ?

drek231 [11]

Hello!

Since we already see an equation that equals y (-2x-4), we can plug that equation into the first to solve for x.

x+4(-2x-4)=19
x-8x-16=19
-7x=35
x=-5

Now that we know the value of x, we will plug it into the second equation to solve for y.

y=-2(-5)-4
y=10=4
y=6

This gives us the answer below

$\left \{ {{x=-5} \atop {y=6}} \right.$

The correct answer is A) (-5,6)

I hope this helps!

7 0

2 years ago

0.5b+4=2(b+2)<br> I need help on this question?

mafiozo [28]

Answer: b=0

Step-by-step explanation:

0.5b+4=2(b+2)

0.5b+4=2b+4

4=1.5b+4

0=1.5b

b=0

8 0

3 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 $a_i=0$ .