This equation has some nested grouping symbols on the left-hand side. As usual, I'll simplify from the inside out. I'll start by inserting the "understood" 1 in front of that innermost set of parentheses:
3 + 2[4x – (4 + 3x)] = –1
3 + 2[4x – 1(4 + 3x)] = –1
3 + 2[4x – 1(4) – 1(3x)] = –1
3 + 2[4x – 4 – 3x] = –1
3 + 2[1x – 4] = –1
3 + 2[1x] + 2[–4] = –1
3 + 2x – 8 = –1
2x + 3 – 8 = –1
2x – 5 = –1
2x – 5 + 5 = –1 + 5
2x = 4
x = 2
It is not required that you write out this many steps; once you get comfortable with the process, you'll probably do a lot of this in your head. But until you reach that comfort zone, you should write things out at least this clearly and completely.
Always remember, by the way, that you can check your answers in "solving" problems by plugging the numerical answer back in to the original equation. In this case, the variable is only in terms on the left-hand side (LHS) of the equation; my "check" (that is, my evaluation at the solution value) looks like this:
LHS: 3 + 2[4x – (4 + 3x)]:
3 + 2[4(2) – (4 + 3(2))]
3 + 2[8 – (4 + 6)]
3 + 2[8 – (10)]
3 + 2[–2]
3 – 4
–1
Since this is what I was supposed to get for the right-hand side (that is, I've shown that the LHS is equal to the RHS), my solution value was correct.
3^3 + 2^7
Step-by-step explanation:
There really is no other way that I can think of other than guessing and checking. 3^3 is 27, when subtracted from 155 gives you 128. 128 can be rewritten as 2^7, leaving your final answer as 3^3 +2^7.
"taken from MrScienceguy"
Answer:
C
Step-by-step explanation:
Fractions cant be integers
The graph is touching the x-axis at x= 2 and is turning away.
When a graph exhibit such a behavior at a root, we say that the multiplicity of the root is 2. So the roots of the polynomial are x = 2 and x = 2. The factors of the polynomial are (x - 2) and (x - 2)
The polynomial thus can be written as product of its factors as (x-2)(x-2)
So the correct answer to this question is option B
Hello,
I don't know how to do this! (if you know , say it.
I have drawn the curves for k=4 to 9
and we only have 3 solutions if 4<k<8
