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4 years ago

50 POINTS!! + BRAINLIEST Find the measure of ∠AED

Mathematics

2 answers:

Eddi Din [679]4 years ago

7 0

Angle AED is supplementary to angle AEB.
Angles AEB and DEC are vertical angles, so their measures are equal.

5x = 3x + 10

2x = 10

x = 5

m<AEB = 5x = 5 * 5 = 25

Angle AED is supplementary to <AEB, so

m<AED + m<AEB = 180

m<AED + 25 = 180

m<AED = 155

Answer is D.

Gnoma [55]4 years ago

3 0

The answer is A.) 25

I just did 3x+10=5x and solved it

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AnnyKZ [126]

Answer:

- 9/2 x 26/3 = -39

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3 years ago

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Sheri's brother gave Sam his collection of stamps when she left for college. At that

pychu [463]

Answer: $70 per year.

Step-by-step explanation:

Let's say that x is the number of years that has passed and y is how much the stamp is worth.

So we know that in zero years the stamp was worth $420 because that is the time Sheri gave her brother Sam the stamp. That could bring up the coordinates (0,420) .

Now we know that in 8 years it was worth $980 and that could be the coordinates (8,980)

To find the rate of change we need to find the different between the y value and divide it by the difference in the x values.

420 - 980 = -560

0-8 = -8

-560/-8 = 70

The rate of change is 70 which means that it grew by $70 every year.

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4 years ago

3x-3y= 4 -3x+3y= 3 how do i solve this equation using the elimination method?

stira [4]

Step-by-step explanation:

3x-3y=4

3y= 3x -4

y= x-4/3

-3x+3y=3

3y= 3x+3

y= x +1

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Since they have the same slope and different y-intercepts , then there are no solutions for these equations because they're parallel.

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3 years ago

Hannah constructed two similar circles as seen in the image below. If the length of the smaller circle is 5/3 pi, which of the f

Bingel [31]

The statement or the choice that states the truth about the two constructed similar circles would be B. The arch length of the larger circle is proportional to the radius of the larger circle. I hope you are satisfied with my answer and feel free to ask for more

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4 years ago

Can someone help me out ! got stuck in this question for an hour.

Reil [10]

Answer:

See below

Step-by-step explanation:

Considering $\vec{u}, \vec{v}, \vec{w} \in V^3 \lambda \in \mathbb{R}$ , then

$\Vert \vec{u} \cdot \vec{v}\Vert \leq \Vert\vec{u}\Vert \Vert\vec{v}\Vert$ we have $(\vec{u} \cdot \vec{v})^2 \leq (\vec{u} \cdot \vec{u})(\vec{v} \cdot \vec{v}) \quad$$

This is the Cauchy–Schwarz Inequality, therefore

$\left(\sum_{i=1}^{n} u_i v_i \right)^2 \leq \left(\sum_{i=1}^{n} u_i \right)^2 \left(\sum_{i=1}^{n} v_i \right)^2$

We have the equation

$\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b} = \dfrac{1}{a+b}, a,b\in\mathbb{N}$

We can use the Cauchy–Schwarz Inequality because $a$ and $b$ are greater than 0. In fact, $a>0 \wedge b>0 \implies ab>0$ . Using the Cauchy–Schwarz Inequality, we have

$\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b} =\dfrac{(\sin^2 x)^2}{a}+\dfrac{(\cos^2 x)}{b}\geq \dfrac{(\sin^2 x+\cos^2 x)^2}{a+b} = \dfrac{1}{a+b}$

and the equation holds for

$\dfrac{\sin^2{x}}{a}=\dfrac{\cos^2{x}}{b}=\dfrac{1}{a+b}$

$\implies\quad \sin^2 x = \dfrac{a}{a+b} \text{ and }\cos^2 x = \dfrac{b}{a+b}$

Therefore, once we can write

$\sin^2 x = \dfrac{a}{a+b} \implies \sin^{4n}x = \dfrac{a^{2n}}{(a+b)^{2n}} \implies\dfrac{\sin^{4n}x }{a^{2n-1}} = \dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}$

It is the same thing for cosine, thus

$\cos^2 x = \dfrac{b}{a+b} \implies \dfrac{\cos^{4n}x }{b^{2n-1}} = \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}}$

Once

$\dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}+ \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}} =\dfrac{a^{2n}}{(a+b)^{2n} \cdot \dfrac{a^{2n}}{a} } + \dfrac{b^{2n}}{(a+b)^{2n}\cdot \dfrac{b^{2n}}{b} }$

$=\dfrac{1}{(a+b)^{2n} \cdot \dfrac{1}{a} } + \dfrac{1}{(a+b)^{2n}\cdot \dfrac{1}{b} } = \dfrac{a}{(a+b)^{2n} } + \dfrac{b}{(a+b)^{2n} } = \dfrac{a+b}{(a+b)^{2n} }$

dividing both numerator and denominator by $(a+b)$ , we get

$\dfrac{a+b}{(a+b)^{2n} } = \dfrac{1}{(a+b)^{2n-1} }$

Therefore, it is proved that

$\dfrac{\sin ^{4n} x }{a^{2n-1}} + \dfrac{\cos^{4n} x }{b^{2n-1}} = \dfrac{1}{(a+b)^{2n-1}}, a,b\in\mathbb{N}$

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3 years ago