I need help with part b and c

Mathematics

1 answer:

Vikki [24]3 years ago

5 0

I am not sure about b cut I think C is 10

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Find the equation of the line passing through (2,-1) and parallel to the line 2x-y=4

Free_Kalibri [48]

Answer:

y = 2x - 5

Step-by-step explanation:

The equation has to have the same slope -- in this case, it is 2 -- to be parallel to 2x - y = 4.

Now we have y = 2x - k.

Then, we plug in the coordinates to our new equation to solve for k.

-1 = 2(2)-k

-1 = 4-k

k = 5

So the final equation would be y = 2x - 5

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3 years ago

A 5 inch tall bamboo shoot doubles in height every 3 days. If the equation y=ab exponent X,where x is the number of doubling per

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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 $k$ :