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Nesterboy [21]
3 years ago
10

4. Twelve nails have a mass of 423 grams. The number of grams per nail is between what 2

Mathematics
1 answer:
salantis [7]3 years ago
4 0

Answer:

C

Step-by-step explanation:

you do twelve divided by 423 to get your answer.

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Please helppp! I'll give brailiest
OLga [1]

Answer:

range of given function is -5/3,-1,4/3

6 0
3 years ago
Item 4 Find the perimeter of the window. Round your answer to the nearest tenth. 90 cm
jeyben [28]

Answer:

Step-by-step explanation:

item 4 Find the perimeter of the window. Round your answer to the nearest tenth. 90 cm

You did not state the diameter or radius of the window

so, i will use 90cm as the diameter of the window

First, I'd calculate the half circumference of the window.

The formula for circumference is c = 2\pi r,

Half of it would just be c/2 = r\pi

Since our diameter is 90 cm, our radius is 45 cm.

Circumference =

=45\pi  \\= 141.37 cm

circumference is also the same with perimeter

<h3>so, the answer is 141.37cm</h3><h3>to nearest tenth = 141.4cm</h3>
5 0
4 years ago
Find the value of angle A and angle B<br> Pls help me with those questions ):
icang [17]

Answer:

5)  A= 153

    B= 27

6)  A= 80

    B= 100

7)  A=88

   B=92

8) A= 30

   B= 150

Step-by-step explanation:

if you look closely all the values are equal to those opposite of them

sorry if its wrong

5 0
3 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
I need help with this question please
ikadub [295]

Answer:

Just connect points Y and D with a straight line to make YD. Do the same for YE and YF, just attach Y to points E and F with a straight line.

5 0
3 years ago
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