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ollegr [7]
2 years ago
13

Z^2 -6z+9=0 Factor and solve

Mathematics
1 answer:
Ymorist [56]2 years ago
8 0
Factor because if we have
yz=0 then we can assume y and/or z=0

so factor
so the easiest way is to find what 2 numbers add to -6 and multiply to get 9 so
we know that they must be both - because we must have x+x=(-) and x^2=+ so therefor they must be negative
factors of 9=3 and 3 so
-3 and -3
-3+-3=-6 perfect

(z-3)(z-3)=0

set each to zero
z-3=0
add 3
z=3


z=3 is the answer
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Which property states that for all real numbers, x, y and z, if x = y and y = z, then x = z ?
Leto [7]

Answer:

The transitive property of equality

Step-by-step explanation:

The transitive property states if a=b and b=c then a=c. If two things are equal and one of those two things is equal to another thing, then they all must be equal.


4 0
3 years ago
Multiple the binomials (x+3)^2
maxonik [38]

Answer:

=x^2+6x+9

Step-by-step explanation:

\left(x+3\right)^2\\\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a+b\right)^2=a^2+2ab+b^2\\a=x,\:\:b=3\\=x^2+2x\cdot \:3+3^2\\\mathrm{Simplify}\:x^2+2x\cdot \:3+3^2:\quad x^2+6x+9\\x^2+2x\cdot \:3+3^2\\\mathrm{Multiply\:the\:num\\bers:}\:2\cdot \:3=6\\=x^2+6x+3^2\\3^2=9\\=x^2+6x+9

8 0
3 years ago
Need help ASAP!!! Please help
cestrela7 [59]
The range of a function is the set of numbers used as y-coordinates.

Let's look at the y-coordinates of the points in the graph.
There are points in the graph with all y-coordinates greater than -3.
There are points in the graph with all y-coordinates less then -3.
At exactly -3, there is no point on the graph with that y-coordinate.
We expect the graph to behave as it is shown above 7 and below -7, so the only value excluded from the y-coordinates is -3.

Answer: B. All real numbers except -3
8 0
3 years ago
Will give brainlyest if u help
goblinko [34]

Answer:

No, because it fails the vertical line test ⇒ B

Step-by-step explanation:

To check if the graph represents a function or not, use the vertical line test

<em>Vertical line test:</em> <em>Draw a vertical line to cuts the graph in different positions, </em>

  • <em>if the line cuts the graph at just </em><em>one point in all positions</em><em>, then the graph </em><em>represents a function</em>
  • <em>if the line cuts the graph at </em><em>more than one point</em><em> </em><em>in any position</em><em>, then the graph </em><em>does not represent a function </em>

In the given figure

→ Draw vertical line passes through points 2, 6, 7 to cuts the graph

∵ The vertical line at x = 2 cuts the graph at two points

∵ The vertical line at x = 6 cuts the graph at two points

∵ The vertical line at x = 7 cuts the graph at one point

→ That means the vertical line cuts the graph at more than 1 point

   in some positions

∴ The graph does not represent a function because it fails the vertical

   line test

7 0
3 years ago
What is the simplified form of<br> (4ab)^2
Alex
Use\ (a\cdot b)^n=a^nb^n\\-----------------\\\\(4ab)^2=4^2a^2b^2=\huge\boxed{16a^2b^2}
7 0
3 years ago
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