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oksian1 [2.3K]
3 years ago
11

Which expression is equivalent to 56÷7?

Mathematics
1 answer:
Nady [450]3 years ago
8 0
Is this a multiple choice question?
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Help me in major bed of some help
zzz [600]
It would be on both planes. When two planes intersect, they form a line along the points at which the two exist in the same place. Point V is said to exist along line s, which came from the intersection of the planes. This means that the point belongs to both of them.
6 0
3 years ago
Data plays a crucial role in decision-making because it can reveal relationships between different quantities. We often use line
mote1985 [20]

Answer:

(a×b)×c = 0 means that vector c parallel to vector (a×b). But because   (a × b) ⊥ a and   (a -b) ⊥ b so

c ⊥ a and c ⊥ b.

Hope this helps, have a nice day! :D

5 0
3 years ago
Read 2 more answers
Given points J(1, 4), A(3,5), and G(2, 1), what arethe coordinates of AJ'A'G' when reflected overthe x-axis?
lilavasa [31]

Reflection across the x-axis:

undefined

y = − f ( x ) y = -f(x) y=−f(x) The concept behind the reflections about the x-axis is basically the same as the reflections about the y-axis.

7 0
8 months ago
Pls help will give brainliest
lana [24]

Answer: The answer is B

Step-by-step explanation: 122 >7 (z+8) -5 (6+z). Then switch them, 7 (z+8) -5 (6+z) <122. Expanded:  7 (z+8) -5 (6+z) <122:v 2z+26

2z+26<122, subtract 26 from both sides: 2z+26-26<122-26

Simplify: 2z<96, divide both sides by 2: 2z over 2 < 96 over 2

and finally simplified: Z<48

HOPE THIS HELPED YOU OUT!

IF NOT SORRY!

6 0
2 years ago
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In e
Sedaia [141]

The probability that, at the tip of the fourth round, each of the players has four coins is 5/192.

Given that game consists of 4 rounds and every round, four balls are placed in an urn one green, one red, and two white.

It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.

Also, let \%R_{A} be the quantity of nonzero elements in R_{A}.

Let C_{1}=\left(\begin{array}{l}1\\ -1\\ 0\\ 0\end{array}\right).

Parity demands that \%R_{A} and\%R_{B} must equal 2 or 4.

Case 1: \%R_{A}=4 and \%R_B=4. There are \left(\begin{array}{l}3\\ 2\end{array}\right)=3 ways to put 2-1's in R_A, so there are 3 ways.

Case 2: \%R_{A}=2 and \%R_B=4. There are 3 ways to position the -1 in R_A, 2 ways to put the remaining -1 in R_B (just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of \%R_{A}=4,\%R_{B}=2 for a complete of 24 ways.

Case 3: \%R_A=\%R_B=2. There are 3 ways to put the -1 in R_{A}. Now, there are two cases on what happens next.

  • The 1 in R_B goes directly under the -1 inR_A. There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in R_C andR_D. (Either the 1 comes first inR_C or the 1 comes first in R_D.)
  • The 1 in R_B doesn't go directly under the -1 in R_A. There are 2 ways to put the 1, and a couple of ways to try and do the identical permutation as within the above case.

Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192

Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.

Learn more about probability and combination is brainly.com/question/3435109

#SPJ4

3 0
2 years ago
Read 2 more answers
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