A rectangle has a height of 2y^3+5 and a width of y^3+6y.

Mathematics

2 answers:

kherson [118]3 years ago

7 0

Answer:

Step-by-step explanation:

vova2212 [387]3 years ago

4 0

Answer:

in polynomial form

Step-by-step explanation:

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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.

7 0

2 years ago

What’s the distance formula

Whitepunk [10]

Step-by-step explanation:

$\text{Let}\ A(x_1,\ y_1)\ \text{and}\ B(x_2,\ y_2).\ \text{Then the distance between}\ A\ \text{and}\ B\ \text{is:}\\\\|AB|=\sqrt{(x_2-x_1)^2+(y_2-y1)^2}\\\\Example:\\\\A(2,\ 5),\ B(-1,\ 1)\\\\|AB|=\sqrt{(-1-2)^2+(1-5)^2}=\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5$