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3 years ago

PLEASE HELP HELPPPPPPPPP

Mathematics

2 answers:

professor190 [17]3 years ago

5 0

Answer:

x+2y=- 7

x-2y =9

Step-by-step explanation:

Solving Steps

2x+4y=-14

x-2y=9

Divide both sides

(2x+4y) ÷2=-14=2

{x-2y=9

Remove the parentheses

Divide

2x÷2+4y÷2=-7

x-2y =9

Divide

Solution

x+2y=- 7

x-2y =9

Scorpion4ik [409]3 years ago

4 0

Answer:

(1,-4)

Step-by-step explanation:

2x+4y=-14

x-2y=9

2x+4y=-14

2(x-2y)=(9)2

2x+4y=-14

2x-4y=18

Add X

4x=4

X=1

Substitute x into any equation

(1)-2y=9

-2y=8

y=-4

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Can y’all help please

IRISSAK [1]

Answer:

x=square root 63. Basically the radical 63.

Step-by-step explanation:

So in the first triangle with the 3 in it. You have root 3 and use the a^2 +b^2=c^2 formula. So 3^2 + 3^2= x^2 and then when you solve you get radical 18^2 which is equal to radical 18 or square root 18. Then we go to the bigger triangle and we do the same formula but this time we have the c value which is 9. So we do radical 18^2 + x^2 = 9^2 and then solve that and minus 18 from the 81 since the radical goes off the 18 and you get square root63.

Sorry i couldn't use the signs. Sorry If I'm wrong Hope this helps!.

4 0

3 years ago

21,985,233 rounded to the millions place

katrin [286]

21,985,233 to the millions place is 22,000,000

6 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 $a_i=0$ .