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Luden [163]
3 years ago
8

Can you help me with these? I need help on how to solve these questions! URGENT!

Mathematics
2 answers:
zhannawk [14.2K]3 years ago
5 0
You can fill the choices into the equations for x and solve then to see if they are correct. For example, you could put -3 and 3/2 into x^2-12x+27=0
Which would make (-3)^2-12(3/2)+27=0.
From the picture I posted, you can see that 0=0, so the equation is correct

adell [148]3 years ago
5 0

Answer:

12. B

13. E

14. D

15. H

16. A

17. F

Step-by-step explanation:

12. (x-3)(x-9)

13. (x+3)(x+9)

14. (x+7)(x-11)

15. (2x+3)(x-3)

16. (2x-3)(x+3)

17. (3x-1)(x-3)

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Does the graph represent a function? Pls help
serious [3.7K]

Answer:

Yes

Basically a scatter plot is a type of function.

5 0
4 years ago
How do you factor 5 out of 5x+40
VARVARA [1.3K]

Answer:

To factor something out you've got to divide, so you would take 5x+40 and divide it all by 5

So, 5x/5= 1x

And, 40/5= 8

Your answer is now x+8 when factored

4 0
3 years ago
Read 2 more answers
Determine the value of z for the system of equations.
Natali5045456 [20]

Answer:

{x, y, z} = {-4, 2, 4}

<u>i think this is the answer</u>

7 0
3 years ago
Can Fractions and ratios cannot have zero in the denominator?
labwork [276]
True , is the answer ! :)
4 0
3 years ago
Which statements are true about the graph of the function f(x) = x2 – 8x + 5? Check all that apply.
statuscvo [17]

Answer:

A, D, E are true

Step-by-step explanation:

You have to complete the square to prove A.  Do this by first setting the function equal to 0, then moving the 5 to the other side.

x^2-8x=-5

Now we can complete the square.  Take half the linear term, square it, and add it to both sides.  Our linear term is 8 (from the -8x).  Half of 8 is 4, and 4 squared is 16.  So we add 16 to both sides.

(x^2-8x+16)=-5+16

We will do the addition on the right, no big deal.  On the left, however, what we have done in the process of completing the square is to create a perfect square binomial, which gives us the h coordinate of the vertex.  We will rewrite with that perfect square on the left and the addition done on the right,

(x-4)^2=11

Now we will move the 11 back over, which gives us the k coordinate of the vertex.

(x-4)^2-11=y

From this you can see that A is correct.

Also we can see that the vertex of this parabola is (4, -11), which is why B is NOT correct.

The axis of symmetry is also found in the h value.  This is, by definition, a positive x-squared parabola (opens upwards), so its axis of symmetry will be an "x = " equation.  In the case of this type of parabola, that "x = " will always be equal to the h value.  So the axis of symmetry is

x = 4, which is why C is NOT correct, either.

We can find the y-intercept of the function by going back to the standard form of the parabola (NOT the vertex form we found by completing the square) and sub in a 0 for x.  When we do that, and then solve for y, we find that when x = 0, y = 5.  So the y-intercept is (0, 5).

From this you can see that D is also correct.

To determine if the parabola has real solutions (meaning it will go through the x-axis twice), you can plug it into the quadratic formula to find these values of x.  I just plugged the formula into my graphing calculator and graphed it to see that it did, indeed, go through the x-axis twice.  Just so you know, the values of x where the function go through are (.6833752, 0) and (7.3166248, 0).  That's why you need the quadratic formula to find these values.

7 0
4 years ago
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