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kotykmax [81]
3 years ago
10

Help me with this please

Mathematics
1 answer:
zheka24 [161]3 years ago
6 0
Three I think, I’m not sure tho so I would get another opinion
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How many<br> roots does the relation y = 4x² - 20x + 25 have? (2 marks)
Dovator [93]

Answer:

it has one root x=2/5

Step-by-step explanation:

7 0
2 years ago
A car travels at an average speed of 52 miles per hour. How long does it take to travel 299 miles
drek231 [11]
The answer should be 299/52=almost 6 hours.
6 0
3 years ago
Which polygon appears to be regular? Figure A Figure B Figure C Figure D Figure A is a hexagon with varying lengths. Figure B is
Naya [18.7K]
Not A because regular means all sides are equal.
B has to be the answer because the others are not.
Not C because a right triangle is not regular, i.e. and equilateral triangle.
Not D because a rectangle has different side lengths.
4 0
3 years ago
Read 2 more answers
"find the reduction formula for the integral" sin^n(18x)
dexar [7]
Let

I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx
For demonstration on how to tackle this sort of problem, I'll only work through the case where n is odd. We can write

\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx
\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx

For the remaining integral, we can integrate by parts, taking

u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax

\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx

For this next integral, we rewrite the integrand

\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax
\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)

So putting everything together, we found

I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx
I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)
I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax
\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax

\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax
7 0
3 years ago
What is the measure of JKL?
Orlov [11]

Answer:

97

Step-by-step explanation:

83 and JKL form a straight line so they add to 180 degrees

83+ JKL = 180

Subtract 180 from each side

83-83+ JKL = 180-83

JKL = 97

7 0
3 years ago
Read 2 more answers
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