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Diano4ka-milaya [45]
3 years ago
15

Two parallel lines, e and f, are crossed by two transversals.

Mathematics
1 answer:
Andrews [41]3 years ago
5 0

9514 1404 393

Answer:

  ∠15 = 97°

Step-by-step explanation:

At any given transversal of parallel lines, all obtuse angles are congruent, and all acute angles are congruent. Obtuse angle 15 is congruent to the one market 97°.

  ∠15 = 97°

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Assume that r varies directly with s. If r = 21 when s = 6, find r when s = 12.
Illusion [34]
\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad  y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------

\bf r=ks\qquad \textit{we also know that }
\begin{cases}
r=21\\
s=6
\end{cases}\implies 21=k6\implies \cfrac{21}{6}=k
\\\\\\
\cfrac{7}{2}=k\qquad therefore\qquad \boxed{r=\cfrac{7}{2}s}
\\\\\\
\textit{if s = 12, what is \underline{r}?}\qquad r=\cfrac{7}{2}\cdot 12
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4 years ago
If your right I’ll mark brainiest need ASAP
Tems11 [23]

Answer:

95 students out of 125 students would be expected to have stleast one sibling

5 0
3 years ago
Read 2 more answers
A restaurant makes cornbread. For each batch of cornbread
serg [7]

Answer:

18

3/4 equals to .75 if you multiply .75 with 21 you get 18

7 0
4 years ago
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 another name for AE ? need asap
polet [3.4K]

Answer:

EA, or s

Step-by-step explanation:

The line AE could also be named any of the following:

AC, EC, CE, CA, EA

If you’re only looking at the line segment AE, the only option is:

EA

Since I don’t know whether you need the line or line segment, you might want go with EA since it’s on both lists. It could also be line s.

7 0
3 years ago
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