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4 years ago

Help please if u can thanks i dont know and understand...

Mathematics

2 answers:

Naily [24]4 years ago

7 0

The line on the left passes through the y axis at x = 3 so it is y = 2x + 3.

b. The line 2x + 1 will be parallel to the other 2 lines because the slope (2 - from the '2x') is the same. It will pass through the y axis at y = 1.

c. The lines -2x + 1 would pass through the y axis at y = 1 but the the slope which is negative 2 will rise to the left unlike the other 3 lines which rise to the right. It will intersect the line y = 2x + 1 at the point (0,1) - where y = 1.

Anon25 [30]4 years ago

6 0

B. the answer is this because i searched it up and got this answer

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Step-by-step explanation:

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3 years ago

Factor the polynomial completely.<br> 2x2 + 8x - 3

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4 years ago

B a C Which of these angles are adjacent angles?

BigorU [14]

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Step-by-step explanation:

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3 years ago

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 $k$ :