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4 years ago

What is the complete factorization of x^2 - 6x + 9

Mathematics

1 answer:

Finger [1]4 years ago

4 0

Answer:

$(x-3)^{2}$

Step-by-step explanation:

Factor using the perfect square rule.

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Inverse square root (x+3)

QveST [7]

Answer:

The inverse function is f(x) = x^2 - 3

Step-by-step explanation:

To find the inverse of any function, start by switching the x and f(x) values.

f(x) = √(x + 3)

x = √(f(x) + 3)

Now solve for the new f(x). The result will be your inverse function.

x = √(f(x) + 3)

x^2 = f(x) + 3

x^2 - 3 = f(x)

8 0

3 years ago

Please answer these it would help me greatly. Thanks!

Leya [2.2K]

Answer:

1. 4in wide

2. 5in tall

3. 10/6in (Equal to 1 4/6in)

4. 1in

5. 1in

I'm pretty sure these answers are right. Hopefully this helped. :)

3 0

3 years ago

Identify the slope and y intercept for each equation y=3x+2

just olya [345]

Answer: Slope = 3

Y-intercept = 2

Step-by-step explanation:

The formula $y=mx+b\\$ is a formula that uses slope and Y-intercept to identify a line on a graph.

m = Slope

b = Y-intercept.

7 0

3 years ago

What is the standard form of 5x^2+5y^2-10x+20y-30=0

o-na [289]

$5x^2+5y^2-10x+20y-30=0$

We need to get the equation in square form

We apply completing the square method

$5x^2-10x+5y^2+20y-30=0$

Divide all the terms by 5

$x^2-2x+y^2+4y-6=0$

Group x terms and y terms separately

$(x^2-2x)+(y^2+4y)-6=0$

By completing the square method, we take half of coefficient of x and square it

$\frac{-2}{2} = -1$

square it $(-1)^2= 1$

Add 1 on both sides

$(x^2-2x+1)+(y^2+4y)-6=+1$

$\frac{4}{2} = 2$

square it $(2)^2= 4$

Add 4 on both sides

$(x^2-2x+1)+(y^2+4y+4)-6=0+1+4$

Add 6 on both sides

$(x^2-2x+1)+(y^2+4y+4)=6+1+4$

Now factor both parenthesis

$(x-1)^2+(y+2)^2=11$

The equation of the circle is $(x-1)^2+(y+2)^2=11$

7 0

4 years ago

Use piece wise function to answer f(0)=

MrMuchimi

Answer:

$f(0) = 0$

Step-by-step explanation:

Given

See attachment

Required

Determine f(0)

To solve for f(0), we make use of:

$f(t) = \frac{t^3}{t + 9}$

Because: $t \ge -2$ is true for $t =0$

So, we have:

$f(t) = \frac{t^3}{t + 9}$

$f(0) = \frac{0^3}{0 + 9}$

$f(0) = \frac{0}{9}$

$f(0) = 0$

7 0

3 years ago