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3 years ago

6

N^2-n-42=0 Please help ASAP !!!!

1 answer:

KatRina [158]3 years ago

8 0

Answer:

(n-7) (n+6)

Step-by-step explanation:

1. You need to find two numbers that's product will equal to 42 and it's sum is -1.

2. You could probably figure out it's 7 & 6 by now and now you have to figure out which one is positive or negative since they both can't be positive or negative since the product needs to be negative and if it will come out to -1. It's simple math from -7 + 6 equals -1. So the 7 would be negative in the equation and the 6 would be positive.

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Answer:

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Step-by-step explanation:

20=20

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4 years ago

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3 years ago

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A steady wind blows a kite due west. The kite's height above ground from horizontal position x = 0 to x = 80 ft is given by y =

GaryK [48]

Answer:

The distance travelled by the kite is 122.8 ft ( approx )

Step-by-step explanation:

Here, the given function,

$y=150-\frac{1}{40}(x-50)^2$

Differentiating with respect to x,

$y'=-\frac{1}{20}(x-50)$

∵ arc length of a curve is,

$L=\int_{a}^{b} \sqrt{1+y'^2}dx$

Where, y shows the height of the curve for a ≤ x ≤ b,

Thus, the arc length of the given curve is,

$L=\int_{0}^{80} \sqrt{1+(-\frac{1}{20}(x-50)^2}dx$

Put $-\frac{1}{20}(x-50)=tan\theta$

$\implies -dx=-20 sec^2\theta d\theta$

$\implies L=-20\int_{0}^{80} \sqrt{1+tan^2\theta}sec^2\theta d\theta$

$=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta$

$=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta$

By integration by parts,

$=|-\frac{20}{2}(sec \theta tan\theta +ln|sec\theta +tan\theta |) |^{x=80}_{x=0}$

If x = 80, $tan \theta = -\frac{1}{20}(30-50)=\frac{3}{2}$

$sec \theta = \frac{\sqrt{13}}{2}$

$\implies \theta = \frac{1}{20}(0-50)=\frac{5}{2}$

$sec \theta = \frac{\sqrt{29}}{2}$

Thus, the length of the curve is,

$=-10(\frac{\sqrt{13}}{2}(-\frac{3}{2}) +ln|\frac{13}{2}-\frac{3}{2}|) + 10(\frac{5\sqrt{29}}{4} + ln |\frac{29}{2} + \frac{5}{2} |)$

$\approx 122.8\text{ feet}$

8 0

3 years ago