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rewona [7]
3 years ago
12

Help with this ill give brainlest

Mathematics
2 answers:
HACTEHA [7]3 years ago
8 0
Can’t really see the problem re post
Archy [21]3 years ago
5 0
Vertical distance is 10 units
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PLEASE HELP ME PLEASE!
Mamont248 [21]

Answer:

-4 and d

==============

Please give brainliest, I really want to rank up, thank you!

5 0
3 years ago
Read 2 more answers
Which pair of expressions below are equivalent? a. −4−8 and 4+8 b. −4−8 and −4+8 c. −4−8 and −4−(−8) d. −4−8 and −4+(−8)
omeli [17]

Answer:

d. −4−8 and −4+(−8)

Step-by-step explanation:

+ - is equivalent to -

6 0
3 years ago
Read 2 more answers
Which defines a circle
SIZIF [17.4K]
Circle<span> is the locus of points equidistant from a given point, the center of the </span>circle<span>. The common distance from the center of the </span>circle<span> to its points is called radius. Thus a </span>circle<span> is completely </span>defined<span> by its center (O) and radius (R): C(O, R) = O(R) = {x: dist(O, x) = R}.
Easier explanation: </span><span>A </span>circle<span> is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure are equal to one another.</span>
4 0
4 years ago
Choose the equation that represents the graph below:
Virty [35]
It would be 9 I think
4 0
4 years ago
Use De Moivre's theorem to write the complex number in trigonometric form: (cos(3pi/5) + i sin (3pi/5))^3
docker41 [41]

By De Moivre's theorem,

\left(\cos\dfrac{3\pi}5+i\sin\dfrac{3\pi}5\right)^3=\boxed{\cos\dfrac{9\pi}5+i\sin\dfrac{9\pi}5}

We can stop here ...

# # #

... but we can also express these trig ratios in terms of square roots. Let x=\dfrac\pi5 and let c=\cos x. Then recall that

\cos5x=c^5-10c^3\sin^2x+5c\sin^4x

\cos5x=c^5-10c^3(1-c^2)+5c(1-2c^2+c^4)

\cos5x=16c^5-20c^3+5c

On the left, 5x=\pi so that

16c^5-20c^3+5c+1=(c+1)(4c^2-2c-1)^2=0

Since c=\cos\dfrac\pi5\neq1, we're left with

4c^2-2c-1=0\implies c=\dfrac{1+\sqrt5}4

because we know to expect \cos x>0. Then from the Pythagorean identity, and knowing to expect \sin x>0, we get

\sin x=\sqrt{1-c^2}=\sqrt{\dfrac58-\dfrac{\sqrt5}8}

Both \cos and \sin are 2\pi-periodic, so that

\cos\dfrac{9\pi}5=\cos\left(\dfrac{9\pi}5-2\pi\right)=\cos\left(-\dfrac\pi5\right)=\cos\dfrac\pi5

and

\sin\dfrac{9\pi}5=\sin\left(-\dfrac\pi5\right)=-\sin\dfrac\pi5

so that the answer we left in trigonometric form above is equal to

\cos\dfrac\pi5-i\sin\dfrac\pi5=\boxed{\dfrac{1+\sqrt5}4+i\sqrt{\dfrac58+\dfrac{\sqrt5}8}}

7 0
3 years ago
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