Answer:
d
Step-by-step explanation:
The answe of this question is d
Answer:
n + 4 = 6
6
Step-by-step explanation:
Given problem:
4 more than one third of a number n is 6
Unknown:
The equation = ?
Solution of the number = ?
Solution:
let the number = n
So, 4 more than one third of a number;
One third of n =
n
4 more;
n + 4
is 6;
n + 4 = 6 (equation of the expression)
Let us now solve the equation:
n + 4 = 6
n = 6 - 4
n = 2
n = 6
To show that they used the same amount of ribbon you could take the 1 yard and divided into different sized equal groups.
1 yard divided by four equals 1/4 yard for each of four ribbons.
1 yard divided by six equals 1/6 yard for each of six ribbons.
You could have ribbons the size of 1/4 our yard and 1/6 yard, but you would have different amounts of each.
Answer:
it changes by 4
Step-by-step explanation:
Answer:
x = 2
Step-by-step explanation:
These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.
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<h3>Squaring</h3>
The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.
x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true
x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution
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<h3>Substitution</h3>
Another way to solve this is using substitution for one of the radicals. We choose ...

Solutions to this equation are ...
u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution
The value of x is ...
x = u² -2 = 2² -2
x = 2 . . . . the solution to the equation
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<em>Additional comment</em>
Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.