All categories

4 years ago

Determine if JK and LM are parallel ,perpendicular, or neither. J(13,-5) K(2,6) L(-1,-5) M(-4,-2)

Mathematics

1 answer:

barxatty [35]4 years ago

7 0

Answer:

The lines are parallel

Step-by-step explanation:

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

Remember that

If two lines are parallel, then their slopes are the same

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

step 1

Find the slope JK

we have

$J(13,-5),K(2,6)$

substitute the values

$m=\frac{6+5}{2-13}$

$m=\frac{11}{-11}$

$m_J_K=-1$

step 2

Find the slope LM

we have

$L(-1,-5),M(-4,-2)$

substitute the values

$m=\frac{-2+5}{-4+1}$

$m=\frac{3}{-3}$

$m_L_M=-1$

step 3

Compare the slopes

$m_J_K=-1$

$m_L_M=-1$

$m_J_K=m_L_M$

therefore

The lines are parallel

You might be interested in

Solve the inequality. 3m ≤ –9 A.m ≥ –3 B.m ≤ –3 C.m ≥ 6 D.m ≤ –6

marin [14]

$3m \leq -9\ /:3 \ \ \ \Rightarrow\ \ \ m \leq -3\ \ \ \Rightarrow\ \ \ Ans.\ B$

6 0

3 years ago

What is the answer to 2 1/3c= 2 1/10

sweet-ann [11.9K]

Answer:

$21 = 209$

hhhhhhhhhhhhhhhhhhhhhh

3 0

4 years ago

Read 2 more answers

-11 2/3 times (-4 1/5)

Julli [10]

Answer:49

Step-by-step explanation:

Converting each to an improper fraction,

-35/3 * -21/5 = 735/15 = 49

3 0

4 years ago

Y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′′(0) = 0

Snowcat [4.5K]

Answer:

y(t) = 3u₂(t) [ $e^{-2t+4} - e^{-5t + 10)}$ ] - 4u₅(t) [ $e^{-2t+10)} - e^{-5t + 25)}$ ]

Step-by-step explanation:

To find - y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′(0) = 0

Formula used -

L{δ(t − c)} = $e^{-cs}$

L{f''(t) = s²F(s) - sf(0) - f'(0)

L{f'(t) = sF(s) - f(0)

Solution -

By Applying Laplace transform, we get

L{y″+ 5y′ + 6y} = L{3δ(t − 2) − 4δ(t −5)}

⇒L{y''} + 5L{y'} + 6L{y} = 3L{δ(t − 2)} − 4L{δ(t −5)}

⇒s²Y(s) - sy(0) - y'(0) + 5[sY(s) - y(0)] + 6Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒s²Y(s) - 0 - 0 + 5[sY(s) - 0] + 6Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒s²Y(s) + 5sY(s) + 6Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒[s² + 5s + 6] Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒[s² + 3s + 2s + 6] Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒[s(s + 3) + 2(s + 3)] Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒[(s + 2)(s + 3)] Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒Y(s) = $\frac{3e^{-2s} }{(s + 2)(s + 3)} - \frac{4e^{-5s} }{(s + 2)(s + 3)}$

Now,

Let

$\frac{1}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3} \\\frac{1}{(s+2)(s+3)} = \frac{A(s + 3) + B(s+2)}{(s+2)(s+3)}\\1 = As + 3A + Bs + 2B\\1 = (A+B)s + (3A + 2B)$

By Comparing, we get

A + B = 0 and 3A + 2B = 1

⇒A = -B

and

3(-B) + 2B = 1

⇒-B = 1

⇒B = -1

So,

A = 1

∴ we get

$\frac{1}{(s+2)(s+3)} = \frac{1}{s+2} + \frac{-1}{s+3}$

So,

Y(s) = $3e^{-2s}[ \frac{1}{(s + 2)} - \frac{1}{(s + 3)}] - 4e^{-5s}[ \frac{1}{(s + 2)} - \frac{1}{(s + 3)}]$

⇒Y(s) = $3e^{-2s} \frac{1}{(s + 2)} - 3e^{-2s} \frac{1}{(s + 3)} - 4e^{-5s}\frac{1}{(s + 2)} + 4e^{-5s}\frac{1}{(s + 3)}$

By applying inverse Laplace , we get

y(t) = 3u₂(t) [ $e^{-2(t-2)} - e^{-5(t - 2)}$ ] - 4u₅(t) [ $e^{-2(t-5)} - e^{-5(t - 5)}$ ]

⇒y(t) = 3u₂(t) [ $e^{-2t+4} - e^{-5t + 10)}$ ] - 4u₅(t) [ $e^{-2t+10)} - e^{-5t + 25)}$ ]

It is the required solution.

3 0

3 years ago

PLEASE HELP! 20 POINTS!An equation was created for the line of best fit from the actual enrollment data. It was used to predict

lakkis [162]

Answer:

B) No, the equation is not a good fit because the residuals are all far from zero.

Step-by-step explanation:

If you add the residuals, you get -60, which is small compared to the scale, but is still far from zero.

3 0

4 years ago

Read 2 more answers