1) -3x -3 -7x = 17
• start by grouping so add -3x and -7x which gives you -10x
• 10x -3 (+3) = 17 +3 = 20 add 3 on both sides)
• 10x (/10) = 20/ 10 = 2
1) x = 2
2) 10x -6 (+6) = 24 + 6 = 30
• 10x (/10) = 30/ 10 = 3
2) x = 3
3) -4 (+4) -5x = -39 +4 = -35
• -5x (/-5) = -35/ -5 = 7
3) x = 7
4) 3x - 5 > 10
this one i don’t understand, is there a typo maybe ?
5) -5x + x + 16 = -3 you can put a 1 to hold the place of x to make it 1x
• (-5x + 1x) + 16 = -3
• -4x + 16 (-16) = -3 -16 = -19
• -4x (/-4) = -19/ -4 = 4.75 or 4 3/4
5) x = 4.75
Answer:
2
Step-by-step explanation:
The reciprocal of a number n is 
Here n = 0.5, then reciprocal is
= 2
1 mpm or 0.0166666666 repeating mph
<span>I note that this problem starts out with "Which is a factor of ... " This implies that you were given several answer choices. If that's the case, it's unfortunate that you haven't shared them.
I thought I'd try finding roots of this function using synthetic division. See below:
f(x) = 6x^4 – 21x^3 – 4x^2 + 24x – 35
Please use " ^ " to denote exponentiation. Thanks.
Possible zeros of this poly are factors of 35: plus or minus 1, plus or minus 5, plus or minus 7. Use synthetic division; determine whether or not there is a non-zero remainder in each case. If none of these work, form rational divisors from 35 and 6 and try them: 5/6, 7/6, 1/6, etc.
Provided that you have copied down the function
</span>f(x) = 6x^4 – 21x^3 – 4x^2 + 24x – 35 properly, this approach will eventually turn up 1 or 2 zeros of this poly. Obviously it'd be much easier if you'd check out the possible answers given you with this problem.
By graphing this function, I found that the graph crosses the x-axis at 7/2. There is another root.
Using synth. div. to check whether or not 7/2 is a root:
___________________________
7/2 / 6 -21 -4 24 -35
21 0 -14 35
----------- ------------------------------
6 0 -4 10 0
Because the remainder is zero, 7/2 (or 3.5) is a root of the polynomial. Thus, (x-3.5), or (x-7/2), is a factor.
The Correct Answer On Edgen Is
<u>A) Linear pair postulate.</u>
