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3 years ago

PLEASE HELP ANYONE HELP ?!!!!!!!!!!!!!!!)))):::::

Mathematics

2 answers:

stealth61 [152]3 years ago

3 0

Answer:

Step-by-step explanation:

givens: x=4

now try your options:

3(4)+3=15

12+3=15

15=15

The first option is correct, therefore the answer is 3x

GuDViN [60]3 years ago

3 0

Answer: I think it is 3x

Step-by-step explanation: I've done this before.

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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 $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

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

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3 years ago

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Answer:

Step-by-step explanation:

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Step-by-step explanation:

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Complete the solution of the equation. Find the value of y when x equals -15. -x ‒ 8y = -1

valkas [14]

To complete the solution of - x - 8y = -1, we first need to substitute x for -15.
As so:

-(-15) - 8y = -1
15 - 8y = -1

Next, we can subtract 15 from each side to get:

15 - 15 - 8y = -1 - 15
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Now, we can divide by -8 on each side to get the y by itself:

-8y / -8 = -16 / -8
y = -16 / -8
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There we go! <em>y</em> = 2 when x = -15! Hope I could help you out! Have a good one.

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