Math plz help!!!! will mark brainliest!! 20 points!

Mathematics

1 answer:

nasty-shy [4]3 years ago

6 0

Perpendicular lines have slopes that are negative reciprocals to each other. the given line has a slope of -2/3, so the perpendicular line's slope is positive 3/2. Either C or D
Next, plug in x=-2, y=-2 to see which one works, C or D?
D works.

or:
use the given point to find the intercept:
given the point (-2, -2), and m=3/2, the equation in point slope form:
(y-(-2)=3/2[(x-(-2)]
y+2=(3/2)(x+2)
y=(3/2)x+1

Again, D is the correct answer.

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Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :