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Artemon [7]
3 years ago
13

Solve the system by substitution 2x-3y=20 x=-2y+3

Mathematics
1 answer:
AfilCa [17]3 years ago
5 0

Answer:

Hi! The answer to your question is (7,-2)

Step-by-step explanation:

☆*: .。..。.:*☆☆*: .。..。.:*☆☆*: .。..。.:*☆☆*: .。..。.:*☆

☁Brainliest is greatly appreciated!☁

Hope this helps!!

- Brooklynn Deka

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Write the fraction in decimal notation 3 1/100
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3.01 is the decimal notation for 3 1/100. You keep the three in the ones place. 1/100 is a decimal. If you use long division or a calculator, you can see that 1/100 is 0.01.
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Would you measure the length of a jump rope in inches or feet explain
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Feet, its in a larger scale, and jump ropes are long objects.
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5 cups to 8 cups <br> identify the percent
KIM [24]
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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.  

5 0
4 years ago
Will ran four miles on his first day of training. The next day he ran one-third that distance. How far did he ran the second day
motikmotik
4/1 times 1/3 is 4/3 or 1 1/3 miles on the second day of training.
8 0
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Read 2 more answers
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