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slavikrds [6]
2 years ago
13

Solve the following systems of equations. To unlock puzzle three

Mathematics
1 answer:
Veronika [31]2 years ago
7 0

Answer:

Step-by-step explanation:

1,)

3x+y=4 <-Equation 1

2x-y=6 <-Equation 2

Step 1 Eliminating Y Multiply negative both side

-[3x +y = 4 ] -

-3x-y = -4 <-Equation 3

Equation 2 to Equation 3

2x-y=6

-

-3x-y = -4

= 5x = 10

x =2

Substituting X to Equation 1

3x+y=4

3(2) + y =4

6 + y = 4

y = -2

Therefore x =2 and y = -2

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7 0
3 years ago
Coldwell elementary has a goal to collect 1,500 box tops in 6 months. the pta brought 458 box tops within two weeks. write and s
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4 0
3 years ago
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34 kg = 5.3 stones

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3 0
3 years ago
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Eric used a remainder theorem to find the remainder of 2x^3 - 4x^2 - 8 my + 1 divided by x - 3. If he calculated the remainder t
IRISSAK [1]

The remainder is non-zero, so x-3 is not a factor of 2x^3-4x^2-8 (or whatever the given polynomial is supposed to be)

6 0
3 years ago
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NikAS [45]
Because these are two consecutive numbers and the first integer is x, the second integer must be x+1 (it could be x-1 too, but here I'm going to solve the problem with x+1).

The sum of the two integers is 57, so we have:
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The two numbers are 28 and 29.

Hope it helps.
5 0
3 years ago
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