4x−6÷x x=6 Help please!

Mathematics

2 answers:

katrin [286]3 years ago

7 0

4(6)-6/(6) = 23

Answer: 23

Leno4ka [110]3 years ago

4 0

Answer:

Step-by-step explanation:

It appears that you want to find the value of the expression 4x - 6 ÷ x when x = 6. Replacing each instance of x with 6, we get:

4(6) - 6 ÷ (6)

Order of opertions rules require that we perform multiplication and division before addition or subtraction. Thus, our

4(6) - 6 ÷ (6) becomes 24 - (6/6), or 24 - 1, or 23.

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Morgarella [4.7K]

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3 years ago

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What if the answer is wrong

stealth61 [152]

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7 0

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 $k$ :