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mihalych1998 [28]
3 years ago
7

How can the Angle-Angle Similarity Postulate be used to prove the two triangles below are similar? Explain your answer using com

plete sentences, and provide evidence to support your claims.
Mathematics
2 answers:
Irina18 [472]3 years ago
4 0
Angle-angle similarity means that if a triangle has two angles some measure the triangles will be similar.

It is true because if the two angles of the triangles are equal, the third angle should also be equal using the interior angles of a triangle is 180. If the three angles of a triangle are all same measure, then the two triangles should be similar.
kipiarov [429]3 years ago
4 0

Answer:

Angle-angle similarity means that if a triangle has two angles some measure the triangles will be similar.

It is true because if the two angles of the triangles are equal, the third angle should also be equal using the interior angles of a triangle is 180. If the three angles of a triangle are all same measure, then the two triangles should be similar.

Step-by-step explanation:

You might be interested in
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
4 years ago
Write the ratio 3 to 10 in two different ways.
sashaice [31]

Answer:

3/10, 3:10

Step-by-step explanation:

3 to 10 : 3/10, 3:10

5 0
3 years ago
I REALLY NEED HELPM ASAP SOON!!!!
Anestetic [448]

The answer is $30 cuz 5 x 6 is 30.
6 0
4 years ago
Points that don't belong to any line​
svetlana [45]

Answer:

Noncollinear points

Step-by-step explanation:

Collinear points lie on the same line. Noncollinear points are the opposite.

3 0
2 years ago
Math- Please answer a few ASAP! Thank you
Brut [27]

Answer:

I  also found all of  the answers  for that package

Step-by-step explanation:

5 0
4 years ago
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