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bazaltina [42]
2 years ago
14

(5+2g)exp5 for g=-2 help pls

Mathematics
1 answer:
Sidana [21]2 years ago
5 0

The value of the expression when g = -2 is -1

<h3>How to simplify the expression</h3>

Given the expression;

(5+2g)exp5

(5+2g)^5

For g = -2

Let's substitute the value of g in the expression

= ( 5 + 2 ( -2) ) ^5

Expand the bracket

= ( 5 - 4) ^ 5

Find the difference

= (-1) ^5

= -1

Thus, the value of the expression when g = -2 is -1

Learn more about algebraic expressions here:

brainly.com/question/4344214

#SPJ1

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This couch is $100.12 and I had $1,000 how much money do I have now
Zanzabum
1,000-100.12=899.88

you have $899.88

hope this helped
7 0
4 years ago
For the function defined by f(t)=2-t, 0≤t&lt;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
4 years ago
7 + r = 15<br><br>A() 22<br><br>B() 8<br><br>C() 105<br><br>D() 9
guajiro [1.7K]

Answer:

B

Step-by-step explanation:

7+r=15

-7    -7

r = 8

B.8

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3 years ago
What is the equation of a line with a slope of 3 and contains<br> point (4.1)
cricket20 [7]

Answer:

y = 3x - 11

Step-by-step explanation:

y=mx+b

1=3(4)+b

1=12+b

b=-11

5 0
3 years ago
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You have a photo that measures 1 in. wide by 3 in. tall. If you want to enlarge the photo so that the height is 18 in., what wil
kiruha [24]
6in... 3 times 6 is 18, so 1 times 6 is 6
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3 years ago
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