All categories

2 years ago

I will mark Brainliest

Mathematics

1 answer:

tamaranim1 [39]2 years ago

3 0

Answer:

Step-by-step explanation:

$75^{30}\\\\75=3\cdot5\cdot5=3\cdot5^2\\\\75^{30}=(3\cdot5^2)^{30}=3^{30}\cdot(5^2)^{30}=3^{30}\cdot5^{60}$

$45^{45}\cdot15^{15}\\\\45=3\cdot3\cdot5=3^2\cdot5\\15=3\cdot5\\\\45^{45}=(3^2\cdot5)^{45}=(3^2)^{45}\cdot5^{45}=3^{90}\cdot5^{45}\\15^{15}=(3\cdot5)^{15}=3^{15}\cdot5^{15}\\\\45^{45}\cdot15^{15}=3^{90}\cdot5^{45}\cdot3^{15}\cdot5^{15}=3^{90+15}\cdot5^{45+15}=3^{105}\cdot5^{60}\\\\=3^{75+30}\cdot5^{60}=3^{75}\cdot\underbrace{3^{30}\cdot5^{60}}_{75^{30}}=3^{75}\cdot75^{30}$

$\text{therefore it is divisible by}\ 75^{30}$

$Used:\\\\(ab)^n=a^nb^m\\\\(a^n)^m=a^{nm}\\\\a^n\cdot a^m=a^{n+m}$

You might be interested in

Find the Circumference and Area of the circle with a diameter of 3 inches.

Vsevolod [243]

Answer:

Circumference-18.84956, rounded, 18.85

Area- 28.27433, rounded-38.27

4 0

2 years ago

Read 2 more answers

1.6[7-4m]=30 2.[5+6a]-3=8

rusak2 [61]

Answer:

(a,m=(-1499/1812, 1/2)

Step-by-step explanation:

hope this helps!

4 0

3 years ago

What is the volume of the priam? 5cm 8cm 152 cm

sasho [114]

Answer:

6080

Step-by-step explanation:

5x8x152=6080

7 0

3 years ago

Read 2 more answers

The nearest star to the earth is <img src="https://tex.z-dn.net/?f=%20%7B2%20%5Ctimes%20%20%2B%204%7D%5E%7B2%7D%20%20%3D%20%2

dalvyx [7]

The nearest start of the earth is Alpha Centauri and the answers to the math question is X= 4 Square root 2

7 0

2 years ago

For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:

Oksi-84 [34.3K]

The half-range sine series is the expansion for $f(t)$ with the assumption that $f(t)$ is considered to be an odd function over its full range, $-1$ . So for (a), you're essentially finding the full range expansion of the function

$f(t)=\begin{cases}2-t&\text{for }0\le t$

with period 2 so that $f(t)=f(t+2n)$ for $|t|$ and integers $n$ .

Now, since $f(t)$ is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

$f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L$

where

$b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt$

In this case, $L=1$ , so

$b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt$
$b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{4-2(-1)^n}{n\pi}$

The half-range sine series expansion for $f(t)$ is then

$f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t$

which can be further simplified by considering the even/odd cases of $n$ , but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming $f(t)$ is even/symmetric across its full range. In other words, you are finding the full range series expansion for

$f(t)=\begin{cases}2-t&\text{for }0\le t$

Now the sine series expansion vanishes, leaving you with

$f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L$

where

$a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt$

for $n\ge0$ . Again, $L=1$ . You should find that

$a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3$

$a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt$
$a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}$
$a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}$

Here, splitting into even/odd cases actually reduces this further. Notice that when $n$ is even, the expression above simplifies to

$a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0$

while for odd $n$ , you have

$a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}$

So the half-range cosine series expansion would be

$f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t$

Attached are plots of the first few terms of each series overlaid onto plots of $f(t)$ . In the half-range sine series (right), I use $n=10$ terms, and in the half-range cosine series (left), I use $k=2$ or $n=2(2)-1=3$ terms. (It's a bit more difficult to distinguish $f(t)$ from the latter because the cosine series converges so much faster.)

5 0

3 years ago