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siniylev [52]
3 years ago
15

280÷3explicacion paso a paso porfavor​

Mathematics
1 answer:
Mademuasel [1]3 years ago
3 0
Lo siento, esto está en inglés, intente usar el traductor

Step 1:
Start by setting it up with the divisor 3 on the left side and the dividend 280 on the right side like this:

3 ⟌ 2 8 0

Step 2:
The divisor (3) goes into the first digit of the dividend (2), 0 time(s). Therefore, put 0 on top:

0
3 ⟌ 2 8 0

Step 3:
Multiply the divisor by the result in the previous step (3 x 0 = 0) and write that answer below the dividend.

0
3 ⟌ 2 8 0
0

Step 4:
Subtract the result in the previous step from the first digit of the dividend (2 - 0 = 2) and write the answer below.

0
3 ⟌ 2 8 0
- 0
2

Step 5:
Move down the 2nd digit of the dividend (8) like this:

0
3 ⟌ 2 8 0
- 0
2 8

Step 6:
The divisor (3) goes into the bottom number (28), 9 time(s). Therefore, put 9 on top:

0 9
3 ⟌ 2 8 0
- 0
2 8

Step 7:
Multiply the divisor by the result in the previous step (3 x 9 = 27) and write that answer at the bottom:

0 9
3 ⟌ 2 8 0
- 0
2 8
2 7

Step 8:
Subtract the result in the previous step from the number written above it. (28 - 27 = 1) and write the answer at the bottom.

0 9
3 ⟌ 2 8 0
- 0
2 8
- 2 7
1

Step 9:
Move down the last digit of the dividend (0) like this:

0 9
3 ⟌ 2 8 0
- 0
2 8
- 2 7
1 0

Step 10:
The divisor (3) goes into the bottom number (10), 3 time(s). Therefore put 3 on top:

0 9 3
3 ⟌ 2 8 0
- 0
2 8
- 2 7
1 0

Step 11:
Multiply the divisor by the result in the previous step (3 x 3 = 9) and write the answer at the bottom:

0 9 3
3 ⟌ 2 8 0
- 0
2 8
- 2 7
1 0
9

Step 12:
Subtract the result in the previous step from the number written above it. (10 - 9 = 1) and write the answer at the bottom.

0 9 3
3 ⟌ 2 8 0
- 0
2 8
- 2 7
1 0
- 9
1

You are done, because there are no more digits to move down from the dividend.

The answer is the top number and the remainder is the bottom number.

Therefore, the answer to 280 divided by 3 calculated using Long Division is:

93
1 Remainder
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Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:
Papessa [141]

Answer: it is verified that:

* y1 and y2 are solutions to the differential equation,

* c1 + c2t^(1/2) is not a solution.

Step-by-step explanation:

Given the differential equation

yy'' + (y')² = 0

To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.

Now,

y1 = 1

y1' = 0

y'' = 0

So,

y1y1'' + (y1')² = (1)(0) + (0)² = 0

Hence, y1 is a solution.

y2 = t^(1/2)

y2' = (1/2)t^(-1/2)

y2'' = (-1/4)t^(-3/2)

So,

y2y2'' + (y2')² = t^(1/2)×(-1/4)t^(-3/2) + [(1/2)t^(-1/2)]² = (-1/4)t^(-1) + (1/4)t^(-1) = 0

Hence, y2 is a solution.

Now, for some nonzero constants, c1 and c2, suppose c1 + c2t^(1/2) is a solution, then y = c1 + c2t^(1/2) satisfies the differential equation.

Let us differentiate this twice, and verify if it satisfies the differential equation.

y = c1 + c2t^(1/2)

y' = (1/2)c2t^(-1/2)

y'' = (-1/4)c2t(-3/2)

yy'' + (y')² = [c1 + c2t^(1/2)][(-1/4)c2t(-3/2)] + [(1/2)c2t^(-1/2)]²

= (-1/4)c1c2t(-3/2) + (-1/4)(c2)²t(-3/2) + (1/4)(c2)²t^(-1)

= (-1/4)c1c2t(-3/2)

≠ 0

This clearly doesn't satisfy the differential equation, hence, it is not a solution.

6 0
3 years ago
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