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xenn [34]
3 years ago
10

Can anybody help me solve or give me answers?

Mathematics
1 answer:
laiz [17]3 years ago
3 0
For questions 9 through 12, plug in the term you're looking for in for x:
9. y=10(20)-2=298
10. y=10(9)-2=88
11. y=5(2^5)=160
12.y=5(2^7)=640

For questions 13 through 16, plug the values 1 through 5 in for x, one at a time:
13. y=4^1=4, y=4^2=16, y=4^3=64, y=4^4=256, y=4^5=1,024
14. y=(-3)^1=-3, y=(-3)^2=9, y=(-3)^3=-27, y=(-3)^4=81, y=(-3)^5=-243
15. y=-3^1=-3, y=-3^2=-9, y=-3^3=-27, y=-3^4=-81, y=-3^5=-243
16. y=10^1=10, y=10^2=100, y=10^3=1,000, y=10^4=10,000, y=10^5=100,000

For question 17, plug 4 in for n:
17. f(4)=5^4=625
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ElenaW [278]

Answer:

Q 12 roots of the equation

2x^{2} -6+1=0 \\= (x-\frac{\sqrt{10} }{2})(x+\frac{\sqrt{10} }{2})\\

∝ = \frac{\sqrt{10} }{2}

β = -\frac{\sqrt{10} }{2}

no matter if u oppose the root

(i) 2(\frac{\sqrt{10} }{2})(-\frac{\sqrt{10} }{2} )^{2}+2(\frac{\sqrt{10} }{2} )^{2}(-\frac{\sqrt{10} }{2})+2(

(ii)((\frac{\sqrt{10} }{2})^{2} - 3 (\frac{\sqrt{10} }{2})(-\frac{\sqrt{10} }{2}) + ((-\frac{\sqrt{10} }{2})^{2}) = \frac{25}{2}

Q 13  roots of equation

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the roots of the second equation are

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x2 = 1/3(1.443) = 0.481

the equation is

(x+0.231)(x-0.481)=0

x^{2}-\frac{1}{4} x-\frac{1}{9}

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