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Savatey [412]
3 years ago
11

The line L1, is y= 5x+1 The line L2, is 2y-10 +3=0 Show how these 2 lines are parallel

Mathematics
1 answer:
anyanavicka [17]3 years ago
4 0

Answer:

Step-by-step explanation:

L₁: y = 5x + 1

Slope of L₁ = 5

L₂: Did you mean 2y - 10x +3 = 0 ?

2y = 10x - 3

y = 5x - 1.5

Slope of L₂ = 5

L₁ and L₂ have the same slope, so the lines are parallel.

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Which equation is equivalent to x + x - (-3) = 12?
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C.) 2x+3 = 12

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Find the square root by long division of 2304​
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8 0
3 years ago
(K3+2k2-82k-28)/(k+10)
Ne4ueva [31]

So for this, we will be using synthetic division. To set it up, have the equation so that the divisor is -10 (since that is the solution of k + 10 = 0) and the dividend are the coefficients. Our equation will look as such:

<em>(Note that synthetic division can only be used when the divisor is a 1st degree binomial)</em>

  • -10 | 1 + 2 - 82 - 28
  • ---------------------------

Now firstly, drop the 1:

  • -10 | 1 + 2 - 82 - 28
  •       ↓
  • -------------------------
  •        1

Next, you are going to multiply -10 and 1, and then combine the product with 2.

  • -10 | 1 + 2 - 82 - 28
  •       ↓ - 10
  • -------------------------
  •        1 - 8

Next, multiply -10 and -8, then combine the product with -82:

  • -10 | 1 + 2 - 82 - 28
  •       ↓ -10 + 80
  • -------------------------
  •        1 - 8 - 2

Next, multiply -10 and -2, then combine the product with -28:

  • -10 | 1 + 2 - 82 - 28
  •       ↓ -10 + 80 + 20
  • -------------------------
  •        1 - 8 - 2 - 8

Now, since we know that the degree of the dividend is 3, this means that the degree of the quotient is 2. Using this, the first 3 terms are k^2, k, and the constant, or in this case k² - 8k - 2. Now what about the last coefficient -8? Well this is our remainder, and will be written as -8/(k + 10).

<u>Putting it together, the quotient is k^2-8k-2-\frac{8}{k+10}</u>

8 0
3 years ago
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