Answer:
1/9 pints
Step-by-step explanation:
[Given] 2/3 / 6
["Keep, change, flip"] 2/3 * 1/6
[Multiply] 2/18
[Simplify] 1/9
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
Answer:
1,1 or -1,3
Step-by-step explanation:
I think that you assumed that:
2k - 1 = k + 2
However we can't do this, because a square has 4 sides whilst a triangle only has 3 sides.
This means that we can say
4(2k - 1) = 3(k + 2)
We multiply by 4 due to 4 sides of the square, and by 3 due to the 4 sides of the triangle.
Lets expand the brackets, and solve it:
4(2k - 1) = 3(k + 2)
8k -4 = 3k + 6
5k -4 = 6 ( subtract both sides by 3x to collect the x values)
5k = 10 (add both sides by 4 to get the x's alone)
k = 2 (divide both sides by 5 to get what just x is)
So k = 2 units
Answer:
<h3>

</h3>
Step-by-step explanation:
x² - 10x + 25 = 35
Move 35 to the left side of the equation
That's
x² - 10x + 25 - 35 = 0
x ² - 10x - 10 = 0
Using the quadratic formula solve the equation
That's
<h3>

</h3>
From the question
a = 1 , b = - 10 , c = - 10
Substitute the values into the above formula and solve
That's

We have the final answer as
<h3>

</h3>
Hope this helps you