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hodyreva [135]
2 years ago
13

What is the vertex of the parabola defined by the equation

Mathematics
1 answer:
avanturin [10]2 years ago
7 0

Answer: B. (2,2)

Step-by-step explanation:

The vertex of the parabola in the form (x-h)^2 = 4p(y-k) is (h, k).

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I don’t understand this I would like some help plz!!!
Andrew [12]

So basically, these questions want you to find the variable<em> x</em>, which you can do by isolating the variable.


3) x * 1/2 = 1/3 <-- To isolate the x in this case, you want to divide both sides by 1/2, aka multiply by 2.

x = 2/3


4) x * 3 = 2 <-- To isolate x, you want to divide both sides by 3

x = 2/3


5) 1/4 * x = 1/6 <-- To isolate x, you want to divide both sides by 1/4, aka multiply by 4.

x = 4/6 = 2/3


6) 4 1/2 * x = 1 1/2 <-- First, let's make these into improper fractions

9/2 * x = 3/2 <-- To isolate x, you want to divide both sides by 9/2, or multiply both sides by 2/9

x = 6/18 = 2/3

8 0
3 years ago
What is the perimeter of the rectangle shown below?
lesantik [10]
D 20 units

I know this because 7+7=14 and 3+3=6. Then 6+14=20.
6 0
3 years ago
Hi help pls its 15 pts:((
ololo11 [35]

Answer:

(x,y) -> (-x,-y)

Step-by-step explanation:

You need the opposites of both points, not just one of them.

4 0
3 years ago
Equations of exponential Functions
natima [27]

Answer:

D. 1.12

Step-by-step explanation:

I calculated it logically

4 0
3 years ago
HELP ASAP 20 POINTS CAUSE IM DESPERATE ToT
Elenna [48]

1. By de Moivre's theorem,

\left(4\left(\cos\left(\dfrac{7\pi}9\right) + i \sin\left(\dfrac{7\pi}9\right)\right)\right)^3 = 4^3 \left(\cos\left(\dfrac{21\pi}9\right) + i \sin\left(\dfrac{21\pi}9\right)\right) \\\\ ~~~~~~~~ = 64 \left(\cos\left(\dfrac{7\pi}3\right) + i \sin\left(\dfrac{7\pi}3\right)\right) \\\\ ~~~~~~~~ = 64 \left(\cos\left(\dfrac\pi3\right) + i \sin\left(\dfrac\pi3\right)\right) \\\\ ~~~~~~~~ = 64 \left(\dfrac12 + i\dfrac{\sqrt3}2\right) \\\\ ~~~~~~~~ = \boxed{32 + 32\sqrt3\,i}

2. First write the given number in exponential/trigonometric form.

z = 5-5\sqrt3\,i

has modulus

|z| = \sqrt{5^2 + \left(-5\sqrt3\right)^2} = \sqrt{100} = 10

and since it lies in the second quadrant of the complex plane, its argument is

\arg(z) = \pi + \tan^{-1}\left(-\dfrac{5\sqrt3}5\right) = \pi + \tan^{-1}\left(-\sqrt3\right) = \pi - \dfrac\pi3 = \dfrac{2\pi}3

So, we have

z = 5 - 5\sqrt3\,i = 10 e^{i2\pi/3} = 10 \left(\cos\left(\dfrac{2\pi}3\right) + i \sin\left(\dfrac{2\pi}3\right)\right)

Now we apply de Moivre's theorem again, and make sure to account for the multivalued-ness of the exponential function. For k\in\{0,1,2,3,4\}, the fifth roots of z are

z^{1/5} = 10^{1/5} e^{i(2\pi/3 + 2\pi k)/5}

k=0 \implies z^{1/5} = 10^{1/5} e^{i2\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{2\pi}{15}\right) + i \sin\left(\dfrac{2\pi}{15}\right)\right)}

k=1 \implies z^{1/5} = 10^{1/5} e^{i8\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{8\pi}{15}\right) + i \sin\left(\dfrac{8\pi}{15}\right)\right)}

k=2 \implies z^{1/5} = 10^{1/5} e^{i14\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{14\pi}{15}\right) + i \sin\left(\dfrac{14\pi}{15}\right)\right)}

k=3 \implies z^{1/5} = 10^{1/5} e^{i20\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{4\pi}3\right) + i \sin\left(\dfrac{4\pi}3\right)\right)} k=4 \implies z^{1/5} = 10^{1/5} e^{i26\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{26\pi}{15}\right) + i \sin\left(\dfrac{26\pi}{15}\right)\right)}

6 0
2 years ago
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