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2 years ago

What is the solution?2x=6

Mathematics

2 answers:

Kryger [21]2 years ago

8 0

Answer:

x = unknown number

and so i think it would be 2(4)=6

2 x 4 = 6?

Step-by-step explanation:

GarryVolchara [31]2 years ago

4 0

Answer:

$\sf\longmapsto \: x = 3$

Step-by-step explanation:

$\sf\longmapsto \: x = 6 \div 2$

$\sf\longmapsto \: x = 3$

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ozzi

Answer:

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Step-by-step explanation:

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1 year ago

Which expression is not equivalent to 2x^2+10x+12

chubhunter [2.5K]

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Step-by-step explanation:

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2 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 $k$ :