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2 years ago

11

Write an equation in slope y-intercept form A(2,6),m=0

1 answer:

REY [17]2 years ago

3 0

Solve for b:

y = 0x + b

6 = 0 + b

b = 6

The answer is y = 0x + 6

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How do I prove this? Help Thanks.

eimsori [14]

Answer:

Step-by-step explanation:

5 0

2 years ago

100000000000000000000000000000000000000000+53624785326348756348927563482564328965342875634289573264

Arlecino [84]

Answer:

5.3624785e+55

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Simplify the expression.<br> c-6 • 2c7<br> A 2с<br> В Зc-42<br> c3c<br> D2c 42

Aleks04 [339]

The answer is that c3x was the answer

4 0

2 years ago

Whats the roots of the following three problems:

klemol [59]

QUESTION 1

The given function is

$f(x)=\frac{(x-3)(x-1)}{(x-1)(x+2)}$

We simplify to get;

$f(x)=\frac{x-3}{x+2}$

This function is equal to zero when $x-3=0$

$\Rightarrow x=3$

QUESTION 2

The given function is

$f(x)=\frac{5x^2-10x+5}{2x^2-5x+3}$

$f(x)=\frac{5(x^2-2x+1)}{2x^2-5x+3}$

We factor to get;

$f(x)=\frac{5(x-1)^2}{(2x-3)(x-1)}$

$f(x)=\frac{5(x-1)}{(2x-3)}$

This function equals zero when

$5(x-1)=0$

$x=1$

QUESTION 3

The given function is

$f(x)=\frac{x^3-8}{x^2-6x+8}$

We factor to get,

$f(x)=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+4)}$

$f(x)=\frac{x^2+2x+4}{x+4}$

The function equals zero when $x^2+2x+4=0$

$D=b^2-4ac$

$D=2^2-4(1)(4)$

$D=-12$

Hence the equation has no real roots.

The complex roots are

$x=-\sqrt{3}i-1$ or $x=-1+\sqrt{3}i$

4 0

2 years ago

Tamara says that raising the number i to any integer power results in either -1 or 1 as the result, since i^2 = -1. Do you agree

Roman55 [17]

I agree only if you have even powers -- even negative ones.

1/i^2 = 1/-1 = - 1
i^0 also gives 1 So far no problem.

It is when you consider the odd numbers that you don't get 1 or -1
You get either -i or i
i^(4n + 1) = i
i^(4n - 1) = -i

5 0

2 years ago