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goldfiish [28.3K]
2 years ago
10

In a class of 24 students, 7 students have 2 brothers or sisters. 4 students have 1 brother or sister. 3 students do not have an

y brothers or sisters. 6 students have 3 brothers or sisters. The remaining students have 4 brothers or sisters.
Samedy says that the total number of brothers and sisters that the class has is 24, because there will be 24 dots above the number line on the line plot. Use the drop-down menus to complete the statement below.

Is Samedy's statement true or false?
In total, the class has _____ brothers and sisters?
Mathematics
1 answer:
Elanso [62]2 years ago
7 0

Answer:

Samdy's statement is FALSE, total number of brothers and sisters is 52

Step-by-step explanation:

To organize our information we should draw a table, where group the students by how many siblings they have, and then calculate the total number of siblings.

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Pachacha [2.7K]

Answer:

B. Negative

Step-by-step explanation:

We are given that x equals negative and y equals positive.

That means the fraction is negative multiplied by the 7y³ term, which is positive.

Negative times positive is negative, so B is our answer.

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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
I dont want the answer i just need help pls and thank you please view image
Elis [28]

I can help! With the Greatest Common Factor you figure out the greatest number that can go into all numbers. The number here would be the 2. You then divide 2 by all the numbers in the problem.
You would end up getting x^2 + 6x + 9


I use to struggle with this too. Hope this helps! Have a great day!!

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3. Put your straightedge along this line. Draw this line as long as you can.
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The answer is 26 trust me I will be ur friend

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