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Svetllana [295]

4 years ago

7

on every 3rd day Ivan goes to the gym. On every fifth day Gavin goes to the gym. What day will Ivan and Gavin will see each othe

r.

2 answers:

Diano4ka-milaya [45]4 years ago

8 0

Find the least common multiple. It would be 15, so on the 15th day.

Marat540 [252]4 years ago

5 0

They will see each other on the fifteenth day

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Which is the additive inverse of the complex number -8+3i?

lawyer [7]

The additive inverse of a complex z is a complex number $-z,$ so that

$\mathsf{z+(-z)=-z+z=0+0i=0}$

Finding $-z:$

$\mathsf{z=-8+3i}\\\\\\ \mathsf{z+(-z)=0+0i}\\\\\ \mathsf{-8+3i+(-z)=0+0i}\\\\ \mathsf{-z=0+0i+8-3i}\\\\ \mathsf{-z=(0+8)+(0i-3i)}\\\\ \boxed{\begin{array}{c}\mathsf{-z=8-3i} \end{array}}\qquad\checkmark$

If you're having problems understanding this answer, try seeing it through your browser: <span>brainly.com/question/2159647

</span>I hope this helps.

Tags: <em>complex number additive inverse opposite algebra</em>

3 0

3 years ago

За – 6а + 2 = 8а + 20 – 5а

Luda [366]

Answer:

a = -3

Step-by-step explanation:

3a - 6a + 2 = 8a + 20 - 5a

Combine Like Terms

-3a + 2 = 3a + 20

-2 -2

-------------------------------------

-3a = 3a + 18

-3a -3a

-------------------------------------

-6a = 18

---- ----

-6 -6

a = -3

4 0

3 years ago

Please help me i will give brainliest!

omeli [17]

Answer:

10

Step-by-step explanation:

1x2x2.5=5

5/0.5=10

6 0

3 years ago

Read 2 more answers

A number is chosen at random from 1-50. find the probability of selecting composite numbers

Ganezh [65]

.68 or a 68% chance

8 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