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Nezavi [6.7K]
3 years ago
5

Determine if it is true or false. p - 4 = 6: p = 10

Mathematics
1 answer:
Mrac [35]3 years ago
5 0

Answer:

its false i think

Step-by-step explanation:

<em>p</em> - 4 = ?

if <em>p</em> equals 6 then turn that to:

6 - 4 =

so that equals 2

6 - 4 = 2

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DOES ANYONE KNOW THIS PLEASE HELP ME ASAPPP. The radius is 6 m if you can’t see it well enough. Thank you. I’ll give 12 points.
cupoosta [38]
The answer is B i’m pretty sure
4 0
3 years ago
A cube and the net for the cube are shown. <br><br> What is the surface area of this cube?
lozanna [386]
Ok, 6×25=150, since there are 6 sides, the answer is 150cm.
3 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
Which number is not equal to one of the following expressions? 2,500 × 10 250,000 ÷ 100 2,500 ÷ 10
ololo11 [35]

Answer:

2,500 ÷ 10

Step-by-step explanation:

Let's evaluate each expression one after the other and find out what they both gives us, then we'd now pick or select the one that doesn't give the required answer.

For 2,500 × 10

2500*10 = 25000

For 250,000 ÷ 10

250000/100= 25000

For 2,500 ÷ 10

2500/10= 250

So from the look of the results,

The one that gave us the odd result is

2,500 ÷ 10, and that's our answer.

4 0
3 years ago
У
Delicious77 [7]

Answer-

y=-2x+2

It starts at +2 so that is the b

And -2x is how you move so that is the m

That's how I was taught and how I remember it lol hope it helped ik it wasn't the best its 12 am here soo I be tired

5 0
4 years ago
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