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2 years ago

6

What is the value of the expression below when z=6? Z to the 2nd power +3z -3

1 answer:

aleksklad [387]2 years ago

7 0

6^2= 36
3 times 6 = 18
36+18-3 =
Answer is 51
I hope this is correct :)

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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

see i don't mind giving my points away..Can anyone help me solve this though for a brainlest and extra points?

victus00 [196]

Answer:

a= $17\sqrt{2}$

Step-by-step explanation:

As we have 2 of the angles, we can find the third. The measure of the third angle is 45 degrees.

As this is the same measure as the other, that means that the side length will be the same.

Now was have two side lengths of 17 in a right triangle, so we can use the Pythagorean theorem to find a.

Recall that the pythagorean theorem states: $a^2+b^2=c^2$

In this case, a is 17, b is 17, and c is a

Knowing this, we can input our value into this formula and solve for a.

$17^2+17^2=a^2\\\\289+289=a^2\\\\a=\sqrt{578} \\\\a=17\sqrt{2}$

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