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2 years ago

What is (0,5) and (2,14) in slope intercept form

Mathematics

1 answer:

stira [4]2 years ago

6 0

Answer:

y = 9/2x + 5

Step-by-step explanation:

ok first find the slope with equation (x2 - x1 over y2 - y1)

14 - 5 = 9/

2 - 0 2

so 9/2 is your slope so now pick one of the points (let's do 0, 5) and plug that into the slope intercept form (y = mx+b)

5 = 9/2(0)+b

9/2 • 0 = 0 so b = 5

so the answer is y = 9/2x + 5. hope this helps!

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Simplify the complex numbers using de Moivre's Theorem and match them with their solutions.

alexandr1967 [171]

For the numbers in $a+bi$ form, convert to polar form:

$1+i=\sqrt2\dfrac{1+i}{\sqrt2}=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)$

By DeMoivre's theorem,

$(1+i)^5=(\sqrt2)^5\left(\cos\dfrac{5\pi}4+i\sin\dfrac{5\pi}4\right)=4\sqrt2\dfrac{-1-i}{\sqrt2}=-4-4i$

$-1+i=\sqrt2\dfrac{-1+i}{\sqrt2}=\sqrt2\left(\cos\dfrac{3\pi}4+i\sin\dfrac{3\pi}4}\right)$

$\implies(-1+i)^6=(\sqrt2)^6\left(\cos\dfrac{18\pi}4+i\sin\dfrac{18\pi}4\right)=8i$

$\sqrt3+i=2\dfrac{\sqrt3+i}2=2\left(\cos\dfrac\pi6+i\sin\dfrac\pi6\right)$

$\implies2(\sqrt3+i)^{10}=2^{11}\left(\cos\dfrac{10\pi}6+i\sin\dfrac{10\pi}6\right)=2^{11}\dfrac{1-i\sqrt3}2=2^{10}(1-i\sqrt3)$

For the numbers already in polar form, DeMoivre's theorem can be applied directly:

$2\left(\cos20^\circ+i\sin20^\circ\right)^3=2\left(\cos60^\circ+i\sin60^\circ\right)=2\dfrac{1+i\sqrt3}2=1+i\sqrt3$

$2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)^4=2(\cos\pi+i\sin\pi)=-2$

At second glance, I think the 2s in the last two numbers should also be getting raised to the 3rd and 4th powers:

$\left(2(\cos20^\circ+i\sin20^\circ)\right)^3=8\left(\cos60^\circ+i\sin60^\circ\right)=4+4\sqrt3$

$\left(2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)\right)^4=16(\cos\pi+i\sin\pi)=-16$

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2 years ago

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Answer : 4x + 1 3/4 = 10

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<span>The product of a constant factor of six and a factor with the sum of two terms, because 6* the sum of y+3.
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3 years ago

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MariettaO [177]

Answer:

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