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lesya [120]
2 years ago
6

NEED HELP ASAP I can’t find the short leg or hypotenuse

Mathematics
2 answers:
insens350 [35]2 years ago
5 0
Memorize these !! They will irritate you with them till you finish university !!

spin [16.1K]2 years ago
3 0

Answer:

short leg = 4

hypotenuse = 8

Step-by-step explanation:

<u>longer leg </u>

12 = x root 3

find x where you will square both sides

x = square root of 48

in surd form is x = 4 root 3

<u>short leg</u>

therefore x = x

x= 4 root 3

<u>Hypotenuse</u>

Y = 2x

Y = 2( 4 root 3)

Y = 8 root 3

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Look at changes of signs to find this has <span>1 </span> positive zero, <span>1 </span> or <span>3 </span> negative zeros and <span>0 </span> or <span>2 </span> non-Real Complex zeros.

Then do some sums...

Explanation:

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Since there is one change of sign, <span><span>f<span>(x)</span></span> </span> has one positive zero.

<span><span><span>f<span>(−x)</span></span>=−3<span>x4</span>+5<span>x3</span>−<span>x2</span>+8x+4</span> </span>

Since there are three changes of sign <span><span>f<span>(x)</span></span> </span> has between <span>1 </span> and <span>3 </span> negative zeros.

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<span><span>f'<span>(x)</span>=−12<span>x3</span>−15<span>x2</span>−2x−8</span> </span>

Newton's method can be used to find approximate solutions.

Pick an initial approximation <span><span>a0</span> </span>.

Iterate using the formula:

<span><span><span>a<span>i+1</span></span>=<span>ai</span>−<span><span>f<span>(<span>ai</span>)</span></span><span>f'<span>(<span>ai</span>)</span></span></span></span> </span>

Putting this into a spreadsheet and starting with <span><span><span>a0</span>=1</span> </span> and <span><span><span>a0</span>=−2</span> </span>, we find the following approximations within a few steps:

<span><span><span>x≈0.41998457522194</span> </span><span><span>x≈−2.19460208831628</span> </span></span>

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Notice the remainder <span>0.013 </span> of the second division. This indicates that the approximation is not too bad, but it is definitely an approximation.

Check the discriminant of the approximate quotient polynomial:

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Since this is negative, this quadratic has no Real zeros and we can be confident that our original quartic has exactly <span>2 </span> non-Real Complex zeros, <span>1 </span> positive zero and <span>1 </span> negative one.

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