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Arlecino [84]
3 years ago
10

The map shows the location of four places in a school. A coordinate plane is shown. There is a point at 5, 2 labeled Front offic

e. There is a point at negative 6, 4 labeled Gym. There is a point at negative 4, negative 3 labeled Auditorium. There is a point at 6, negative 4 labeled Library. Peter's English class is located in the same quadrant is the auditorium. Which of the following could be the coordinates of the English classroom? (3, 2) (−3, 2) (−3, −2) (3, −2)
Mathematics
2 answers:
SIZIF [17.4K]3 years ago
8 0

Answer:

(-3,2)

Step-by-step explanation:

Molodets [167]3 years ago
3 0

Answer:

-2,5 since it is in the same quadrant of -6,4 which Is the gym

Think about it this way. Imagine a graph in your head. Left is negative. Right is positive. Up is positive and down is negative. Think of the numbers as rise over run. Vertical then horizontal. Because the gym is negative then positive you would need a number that is to the left then up.

Hope this helped:)

Step-by-step explanation:

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What is 3.754×10^5 than round it to the tenth
saul85 [17]

Answer:

375,400

Step-by-step explanation:

  1. 3.754*10^5
  2. 10^5=100,000
  3. 3.754*100,000
  4. 375,400

YaYYYY u got it! =)

5 0
3 years ago
Read 2 more answers
Is x+y=6 a direct variation? Why or why not?
ivanzaharov [21]
I think the variation is x. because for have the y-intercept if yo move x
5 0
3 years ago
HELP PLEASE!!! I need help with 94 if you could show the steps that would be very helpful!
aksik [14]
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:

n!/((n-r)!r!)

In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.

5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5

5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10

5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10

5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5

Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.

Now we add together the combinations

1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.

The answer is 32.

-------------------------------

Note: There is also an equation for permutations which is:

n!/(n-r)!

Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.

We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.

I hope this helps! If you have any questions, let me know :)








7 0
3 years ago
Pls help I’ll brainlest grades close today
exis [7]

Answer:

12.5%

Step-by-step explanation:

[First Value – Second Value] ÷ [(First Value + Second Value) ÷ 2] × 100

[50 - 30] ÷ [ 50 + 30] ÷ 2 × 100

20 ÷ 80 ÷ 2 × 100

0.25 ÷ 2 × 100

0.125 × 100

12.5 %

8 0
3 years ago
Read 2 more answers
1/3(x+3) - 2/x+3, where x is not equal to negative 3
myrzilka [38]
If x\neq 3, then

\frac{1}{3(x+3)} - \frac{2}{x+3} = \frac{1\cdot 1-2\cdot 3}{3(x+3)} = \frac{-5}{3(x+3)} =- \frac{5}{3(x+3)}.

5 0
3 years ago
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