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4 years ago

3(8x-5)=18x+33 =true or false

Mathematics

1 answer:

Tju [1.3M]4 years ago

3 0

Answer:false

Step-by-step explanation:

3(8x-5) = 24x-15

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Please help me answer this question. First to answer will be marked braineist. NO WORK NESSACARY. Thank you ☺️

m_a_m_a [10]

Lets check the first point (0,6)

3*0+4=4 no good

0+6 =6 good

0^2+6 =6 good

6 * (7/6) ^0 =6 good

Lets check the second point (1,7)

3*1+4= we don't care since it didn't work for (0,6)

1+6 =7

1^2+6 =7

6 * (7/6) ^1 =7

Lets check the third point (2,10)

3x+4= we still don't care

2+6 = 8 no good

2^2+6 =10 good

6 * (7/6) ^2 =6 * 49/36 =49/6 no good

f(x) = x^2 +6

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3 years ago

The coordinates of points A and L are -4 and 12, respectively. What is the coordinate of the midpoint between them?

Nostrana [21]

I'm not Totally sure but I'm pretty sure its B<span />

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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 $k$ :