For this case we must algebraically rewrite the given expression. So:
x: It is the variable that represents an unknown number
"double a number" is represented as: 
"6 less than double a number" is represented as: 
So, the final expression is:
If we want to rewrite the expression in an equivalent way, we take common factor 2:
Asnwer:
Equivalent expression:
So we started with 25, then after we lost a solid 4 as a given variable. So now we have 21 beans. Later we find that 6 pounds are left and are in not part of the beans sold that can be effectively used to find the weight of each equal bean bag. So we subtract 6 from 21. We get 15 pounds that were sold in 20 bags. The equation to find the weight of each bag from there is :
15 lb/10 bags
From there it equals 0.75 lb per bag.
Answer:
20x + 15 = 11
Step-by-step explanation:
First you need to add the like terms:
-7+2=-5
Now you need to expand the brackets:
-5*-4x=20x
-5*-3=15
Now you have 20x + 15 = 11. So if you want you can rearrange the formule to find x.
20x=-4
x= -5
Answer:
Step-by-step explanation:
Let as us 45 as the number we should use to get the square root in the simplest form
And then 144
The square root of a number in is simplest form means to get the number inside the radical as low as possible
Answer: Choice B
(x-1)(x^3+x^2+5x+6)
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Explanation:
The 1 in the upper left box means that x = 1 is a root of the original polynomial.
So this means x-1 is a factor of the original polynomial.
This is because x = 1 leads to x-1 = 0 after subtracting 1 from both sides.
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The 0 in the last position of the bottom row shows we got a remainder of 0.
Getting a remainder 0 tells us that (x-1) is a factor of the polynomial. This synthetic division table confirms our initial guess.
The other values in that bottom row (1, 1, 5, 6) form coefficients to the polynomial 1x^3+1x^2+5x+6, or simply x^3+x^2+5x+6
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So we know that (x-1) and (x^3+x^2+5x+6) are factors
Meaning that,
x^4+4x^2+x-6 = (x-1)(x^3+x^2+5x+6)
You can confirm this by expanding out the right hand side (distribution rule).
