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Juliette [100K]
3 years ago
11

Using the common​ denominator of 56, what is an equivalent fraction for 3/8 and 3/7?

Mathematics
1 answer:
anyanavicka [17]3 years ago
5 0

Answer: \frac{3}{8}=\frac{21}{56}   \frac{3}{7}=\frac{24}{56}

Step-by-step explanation:

In order to get from 8 to 56, we have to multiply by 7.

We always do the same to the top! 3 multiplied by 7 = 21.

In order to get from 7 to 56, we have to multiply by 8.

We always do the same to the top! 3 multiplied by 8 = 24.

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I need help with this proof, the given and prove are at the top. Plzzzz help
Goshia [24]

Answer:  The proof is mentioned below.

Step-by-step explanation:

Here, Given : m∠AOC = 160°  m∠AOD= (3x-10)° and  m∠ DOC= (x+14)°

Prove: x= 39°

      Statement                                              Reason

1. m∠AOC = 160°, m∠AOD= (3x-10)°    1.  Given

and m∠ DOC= (x+14)°    

2. m∠AOD+m∠DOC=m∠AOC         2. Because OD divides ∠AOC

                                                              into ∠AOD and ∠DOC

3. (3x-10)° +(x+14)°= 160°                     3. By substitution

4.  4x+4 = 160°                                    4. By equating like terms

5. 4x= 156°                                          5. By subtraction property

                                                             of equality

6. x= 39°                                              6. By division property of equality


7 0
3 years ago
3x + y = 7<br><br> 6x + 2 = 14<br><br> HELP ME GRAPH PLEASEE
MAVERICK [17]

Answer:

Step-by-step explanation:

6 0
2 years ago
Three equivalent ratios of 10 to 15
wariber [46]
10/15 10 to 15 10:15 are answers
5 0
3 years ago
'Polonium-210 is a radioactive substance with a half-life of 138 days. If a nuclear
Tatiana [17]

Answer:

  • <u>59.0891 g (rounded to 4 decimal places)</u>

Explanation:

<em>Half-life time</em> of a radioactive substance is the time for half of the substance to decay.

Thus, the amount of the radioactive substance that remains after a number n of half-lives is given by:

  • A=A_0\cdot (1/2)^n

Where:

  • A is the amount that remains of the substance after n half-lives have elapses, and
  • A₀ is the starting amount of the substance.

In this problem, you have that the half-live for your sample (polonium-210) is 138 days and the number of days elapsed is 330 days. Thus, the number of half-lives elapsed is:

  • 330 days / 138 days = 2.3913

Therefore, the amount of polonium-210 that will be left in 330 days is:

  • A=310{g}\cdot (1/2)^{2.3913}=59.0891g
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
3 years ago
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