If you know that y = 6. You just have to switch Y by 6.

x + 2y = 16
x + 2*6 = 16
x + 12 = 16
x = 16 - 12
x= 4

4 + 2*6 = 16

Hope this helps !

Photon

7 0

3 years ago

1). −3p + 6 p =

Katena32 [7]

1.) 3p
2. -b+3
3. 6x
4. -3p
5. -4v
6. r
7. 9-4r
8. 4b+2
9. 14n
10. 4b+2
11. 35n+45
12. -82v
13. 38n
14. 38x

4 0

3 years ago

Read 2 more answers

If b = 5, what is b to the square root of 3

liubo4ka [24]

Is 125 because u need to take the 5 and add 3 on the top

6 0

3 years ago

When the solutions are going to be complex, which methods can be used to solve quadratic equations?

yKpoI14uk [10]

You could complete the square to state the vertex.
You could use the quadratic equation to find the roots (which are complex).

Try an example that will require both.

y = x^2 + 2x + 5

Step One
Get the graph. That's included below.

Step Two
Provide the steps for completing the square.
Note: we should get (-1,4)

y= (x^2 +2x ) + 5
y = (x^2 +2x + 1) + 5 - 1
y = (x +1)^2 + 4

The vertex is at (-1,4)

Step Three
Find the roots. Use the quadratic equation. Note that the graph shows us that the equation never crosses or touches the x axis. The roots are complex.

$\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a}$

a = 1
b = 2
c = 5

$\text{x = }\dfrac{ -2 \pm \sqrt{2^{2} - 4*1*5 } }{2}$
$\text{x = }\dfrac{ -2 \pm \sqrt{4 - 20 } }{2}$
$\text{x = }\dfrac{ -2 \pm \sqrt{-16 } }{2}$
$\text{x = }\dfrac{ -2 \pm 4i }{2}$
x = -1 +/- 2i

x1 = -1 + 2i
x2 = -1 - 2i And we are done.

7 0

3 years ago