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3 years ago

Please help! Name an angle adjacent and congruent to angle AOC ( diagram in picture)

Mathematics

1 answer:

kipiarov [429]3 years ago

7 0

Answer:

Angle DOC

Step-by-step explanation:

Since angle AOC and angle DOC are ajacent angles that form a straight line, they are supplementary to each other.

By the definition of supplementary angles, you can determine that angle AOC + angle DOC = 180.

Also notice that angle DOC is a right angle (90 degrees), so you can also determine that angle AOC is also a right angle since they are supplementary.

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The vertex of this parabola is at (-3, 6). Which of the following could be its equation?

12345 [234]

Nice, already in vertex form
y=a(x-h)^2+k
(h,k) is vertex

therfor since (-3,6) is vertex
we are looking for something like
y=a(x-(-3))^2+6 simplified to
y=a(x+3)^2+6

A is ansre

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3 years ago

Read 2 more answers

every year, Hannah sells blueberry and strawberry jam at the local farmers market. Hannah charges four dollars Per jar of bluebe

bazaltina [42]

Answer: part A- $4b+6s=c$

part B- $4[3]+6[5]=42$

Step-by-step explanation:

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3 years ago

two bags and four hats cost $100 in all. three bags and seven hats cost $164 in all. what is the cost of 1 hat?

Natali [406]

Answer:

Bags = 22

Hat = 14

22 x 2 = 44

14 x 4 = 56

So the first part is true

3 x 22 = 66

7 x 14 = 98

So the second part is true

14 is the answer

It's trial and error

Step-by-step explanation:

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3 years ago

An American green tree frog tadpole is about 0.00001 kilometer in length when it hatches. Write this decimal as a power of 10

galben [10]

1.e-50
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3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

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3 years ago