We have that
<span>[6x+5]=1+2*(3x+2)
[6x+5]=1+2*3x+2 --------> is not correct ------->1+ 2*(3x+2)=1+2*3x+2*2
then
</span>[6x+5]=1+6x+4---------------> [6x+5]=[6x+5]
<span>this equation is an identity, all real numbers are solutions.</span>
1 and 2 i can't answer without proper data.
3. x+2x+2x+6 = 86
5x+6 = 86
-6 -6
5x = 80
x = 16
Side one: x is 16
Side two: 2x is 2(16) = 32
Side three: 2x+6 is 38
The two containers hold 328 ounces at the they hold same amount of water.
<u>Step-by-step explanation:</u>
The equations below model the ounces of water, y, in each container after x minutes.
At the time after the start when the containers hold the same amount of water, the two equations must be equal.
⇒ 
The first step is to divide everything by 2 to make it simplified.
⇒ 
Now put everything on the left
.
Add the like terms together to further reduce the equation
Factorizing the equation to find the roots of the equation.
Here, b = -12 and c = -28
where,
- b is the sum of the roots ⇒ -14 + 2 = -12
- c is the product of the roots ⇒ -14 × 2 = -28
- Therefore, (x-14) (x+2) = 0
- The solution is x = -2 or x = 14
Take x = 14 and substitute in any of the given two equations,
⇒ 
⇒ 
⇒ 328 ounces
∴ The two containers hold 328 ounces at the they hold same amount of water.
Answer:
Step-by-step explanation:4log(5) = log(5^4) = log(625).
This problem involves using one of the properties of logs, where a coefficient (in this case the "4") for a logarithm equals the "inside of a logarithm" raised to power of whatever number the coefficient is.
The property in mathematical terms is: Alog(B) = log(B^A).
So, 4log(5)= log(5^4) = log(625)
No.
A fifth degree polynomial, having a graph that increases and starts from below x-axis.
Therefore, no matter what equation it is. The fifth degree polynomial will intercept x-axis AT LEAST one.
The fifth degree polynomial can have only at maximum, 4 complex roots.
<em>You can try drawing or seeing the graph of fifth-degree polynomial function. No matter what equations, they still intercept at least one x-value.</em>
<em />
