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3 years ago

15

Is 63% a real number?

1 answer:

Elden [556K]3 years ago

8 0

No I do not believe it is.

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POINTS. POINTS. PLEASE ANSWER CORRECTLY.

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what came first the chicken or the egg

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2 years ago

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What’s the distance formula

Whitepunk [10]

Step-by-step explanation:

$\text{Let}\ A(x_1,\ y_1)\ \text{and}\ B(x_2,\ y_2).\ \text{Then the distance between}\ A\ \text{and}\ B\ \text{is:}\\\\|AB|=\sqrt{(x_2-x_1)^2+(y_2-y1)^2}\\\\Example:\\\\A(2,\ 5),\ B(-1,\ 1)\\\\|AB|=\sqrt{(-1-2)^2+(1-5)^2}=\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5$

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3 years ago

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Can anyone help me factor polynomials

Anarel [89]

C) x+3
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3 years ago

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Vinvika [58]

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Step-by-step explanation:

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3 years ago

Find the first three iterates of the function f(z) = z2 + c with a value of c = 2 - 3i and an initial value of z0 = 1 + 2i.

Artist 52 [7]

Answer:

C

Step-by-step explanation:

We have: (I rewrote the function)

$f(z_n)={z_{n-1}} ^2+c$

Given that:

$\displaystyle c=2-3i \text{ and } z_0 = 1 + 2 i$

The first iterate will be:

$\displaystyle \begin{aligned} f(z_1)&=(z_0)^2+c \\ &=(1+2i)^2+(2-3i) \\ &= (1+4i+4i^2)+(2-3i) \\ &=1+4i-4+2-3i \\ &=-1+i \end{aligned}$

The second iterate will be:

$\begin{aligned}f(z_2) &=(z_1)^2+c\\ &=(-1+i)^2+(2-3i) \\&= (1-2i+i^2)+(2-3i) \\&=1-2i-1+2-3i \\&=2-5i \end{aligned}$

And the third iterate will be:

$\begin{aligned} f(z_3)&=(z_2)^2+c\\ &=(2-5i)^2+(2-3i) \\ &=(4-20i+25i^2)+(2-3i) \\ &=4-20i-25+2-3i \\ &=-19-23i \end{aligned}$

Hence, our answer is C.

3 0

3 years ago