Answer:
THIS IS TOO CONFUSING
Step-by-step explanation:
Your answer is going to be
23% :)
If I'm understanding the question correctly, 7+3=10. 10*8=80.
Answer:
3 dogs and 2 cats.
Step-by-step explanation:
In this case we can solve it using a 2x2 system of equations, like this:
let x: number of cats
let y: number of dogs
So:
42.5 * x + 64 * y = 277
35.5 * x + 50.5 * y = 222.5 => x = (222.5 - 50.5 * y) /35.5
Replacing, we are left with that:
42.5 * (222.5 - 50.5 * y) /35.5 + 64 * y = 277
266.37- 60.45 * y + 64 * y = 277
64 * y - 60.45 * y = 277 - 266.37
3.55 * y = 10.63
y = 10.63 / 3.55
y = 2.99, about 3
Now to calculate x:
x = (222.5 - 50.5 * 3) /35.5
x = 2
Which means that it has a total of 3 dogs and 2 cats.
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.