Help asap!will mark brainliest if there's two answers thank you so much if you help

What is the equation of the line that is parallel to the given line and passes theought the point (12, -2)

Nesterboy [21]

Answer:

Option D (y = 5/6x -12).

Step-by-step explanation:

Hey there!

The equation of the line which passes through the point (12,-2) is (y+2) = m2(x-12)………(i) [Using one point formula].

According to the question, the first line passes through point (12,6) and (0,-4).

So,

$slope(m1) = \frac{y2 - y1}{x2 - x1}$

$or \: m1 = \frac{ - 4 - 6}{0 - 12}$

$or \: m1 = \frac{5}{6}$

Therefore, the slope of the line is 5/6.

Now as per the condition of parallel lines, m1 =m2 = 5/6.

So, keeping the value of m2 in equation (i), we get;

(y+2) = 5/6(x-12)

$y + 2 = \frac{5}{6} x - 10$

or, y = 5/6x - 12.

Therefore, the required equation is y = 5/6 X - 12.

Hope it helps!

3 0

2 years ago

HELP! 40 points!

Vitek1552 [10]

Y-14=12# x+1 answer of math

7 0

2 years ago

Read 2 more answers

Find the value of x. B х 8 A 3 D 4 x = [?] Enter

omeli [17]

Answer:

X=6

Step-by-step explanation:

If B and C equal 8 and is larger than x than x is a smaller number. D and C is 4 but its smaller than x so x has to be bigger. Your left with 3 choices 5, 6, and 7. Measure it and find out how big x is and its between 6 or 7. If any means they do decimals it can also be 6.5 so x=6 or x=7 or x=6.5

4 0

2 years ago

Suppose you are interested in testing whether the mean earning of men in California differs significantly from the mean earning

notsponge [240]

Answer:

74

Step-by-step explanation:

4 0

2 years ago

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago