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2 years ago

Please help don’t give any fake links pls

Mathematics

1 answer:

anygoal [31]2 years ago

8 0

Answer:

x²+y²+4y -5=0

Step-by-step explanation:

x²+(y+2)²=3²

<=> x²+y²+4y+4=9

=> x²+y²+4y -5=0

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There are tickets numbered 1 to 20 that are mixed up and drawn at random. What is the probability that the ticket drawn is numbe

Travka [436]

Answer:

2/20

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,

6 0

3 years ago

Read 2 more answers

What is x 3x-11+9=10

Archy [21]

Answer: x = 4

Step-by-step explanation:

3x - 11 + 9 = 10

Add 11 to each side

3x + 9 = 21

Subtract 9 from each side

3x = 12

Divide each side by 3

x = 4

3 0

3 years ago

Read 2 more answers

What is 2,912 rounded to the nearest thousand

FrozenT [24]

Answer:

3,000

Step-by-step explanation:

Find the number in the thousand place 2 and look one place to the right for the rounding digit 9 . Round up if this number is greater than or equal to 5 and round down if it is less than 5 .

5 0

3 years ago

Read 2 more answers

Which of the following is true about the expression 3 Squared times 12 squared

ololo11 [35]

Answer:

Step-by-step explanation:

the expression would look like this: 3^2 x 12^2

3 squared is 9 (3 x 3)

12 squared is 144 (12 x 12)

3 0

3 years ago

"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