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nikklg [1K]
2 years ago
8

Write an integer for this situation: a profit of $12

Mathematics
1 answer:
finlep [7]2 years ago
3 0

Answer:

+12

Step-by-step explanation:

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Help please !!!!!!!!!!!!!
cupoosta [38]
The answer is 3.....
2x+5
2x3+5=11      2x3=6
4x-1
4x3-1=11       4x3=12
7 0
3 years ago
a circuit overloads at 1800 watts of electricity you plug a microwave oven that uses 1100 watts into the circuit write and solve
Bogdan [553]
Let w be the additional watts you can add on the circuit.

1100 + w < 1800 (write the inequality)
-1100 -1100 ( - 1100 from each side)

w < 700 (simplify)


ANSWER:
You can add up to 700 watts to the circuit, which means that you can also plug in the clock radio and the blender.


LOOK BACK:
You can check that your answer is correct by adding the numbers of watts used by the microwave oven, clock radio and blender.

1100 + 50 + 300 = 1450

The circuit will not overload because the total wattage is less than 1800 watts.
5 0
3 years ago
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What is the greatest common factor of the numbers 12 and 54?
Anuta_ua [19.1K]

Answer:

6

Step-by-step explanation:

54=2x3x3x3

12=2x2x3

common factors are 2 and 3 so 2x3=6

3 0
3 years ago
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Mindy is playing a game. She notices that the ratio of red tokens to purple tokens is 2 to 3. She counts 6 red tokens. Draw numb
Ne4ueva [31]
Do you is playing games notices that the ratio of red tokens to purple tokens is 2 to 3 to count six red tokens John numbers to complete the diagram to the number of purple tokens numbers may be used once more than once or not at all.

Answer:
4 0
3 years ago
Please help with imaginary numbers worksheet!
mihalych1998 [28]

Problem 5

<h3>Answer:   6i</h3>

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

Explanation:

We have these four identities

  • i^0 = 1
  • i^1 = i
  • i^2 = -1
  • i^3 = -i

Notice how computing i^4 leads us back to 1. So i^4 = i^0. The pattern repeats every 4 terms. So we divide the exponent by 4 and look at the remainder. We ignore the quotient entirely. We can see that 28/4 = 7 remainder 0. Meaning that i^28 = i^0 = 1.

We can think of it like this if you wanted

i^28 = (i^4)^7 = 1^7 = 1

Then the sqrt(-36) becomes 6i

So overall, we end up with the final answer of 6i

=============================================

Problem 6

<h3>Answer:  -3i</h3>

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

Explanation:

We'll use the ideas mentioned in problem 5

46/4 = 11 remainder 2

i^46 = i^2 = -1

sqrt(-9) = 3i

The two outside negative signs cancel out, but there's still a negative from -1 we found earlier. So we end up with -3i

In other words, here is one way you could write out the steps

-i^{46}*-\sqrt{-9}\\\\-i^{2}*-3i\\\\i^{2}*3i\\\\-1*3i\\\\-3i\\\\

=============================================

Problem 7

<h3>Answer:   -1</h3>

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

Work Shown:

i^10 = i^2 because 10/4 = 2 remainder 2

i^19 = i^3 because 19/4 = 4 remainder 3

i^7 = i^3 because 7/4 = 1 remainder 3

Again, all we care about are the remainders.

i^{10}+i^{19} - i^{7}\\\\i^{2}+i^{3} - i^{3}\\\\i^{2}\\\\-1

=============================================

Problem 8

<h3>Answer:    -1 + i</h3>

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

Work Shown:

i^22*i^6 = i^(22+6) = i^28

Earlier in problem 5, we found that i^28 = i^0 = 1

So,

i^1 - \left(i^{22}*i^{6}\right)\\\\i^1 - i^{28}\\\\i^1 - i^{0}\\\\i - 1\\\\-1+i

8 0
2 years ago
Read 2 more answers
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