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Gala2k [10]
3 years ago
11

A class has 29 students. In how many different ways can four students form a group for an​ activity? (Assume the order of the st

udents is not​ important.)
Mathematics
1 answer:
loris [4]3 years ago
3 0

Answer:

7.25

Step-by-step explanation:

29 divided by 4

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Brad can make 4 key chains in an hour. Velma can make only 3 key chains in an hour, but she already has 6 completed key chains.
DIA [1.3K]
B=4t
V=3t+6
they have the same amount of key chains when they equal
4t=3t+6
t=6
it would take 6 hours

5 0
3 years ago
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3x+10y=31 <br> 2x + 10= y <br> solve the problem with substitution
barxatty [35]

Answer:

(-3,4) or x = - 3, y = 4

Step-by-step explanation:

3 0
4 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
Sydney makes $12.00 per hour working at GreyStone. Last week he worked 38 hours. What is his gross pay?
telo118 [61]

38 x $12.00 = $456.00

So Sydney's gross pay is A) $456.00.

Hope this Helps!!

7 0
3 years ago
Solve the equation: 3.2(s+10)=32
Lelechka [254]
3.2(s+10)=32

3.2s+20=32

6s+20=32

6s=12

s=2
8 0
3 years ago
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