Answer:
25 Balls
Step-by-step explanation:
<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.
Answer:
The slope of the function graphed is less than the linear function that has an x-intercept of 1 and a y-intercept of -2.
Step-by-step explanation:
It is given that x-intercept is 1 and y-intercept is -2.
Now, the coordinate of the point is (1,0) and (0,-2).
We know that,
Slope = 
Now, Slope of the graph given
Co-ordinate is (-5,0) and (1,2).
Using the formula of slope
Now, 
Hence, the slope of function graphed is less than linear function that has an x-intercept of 1 and a y-intercept of -2.
Area of the square is 144m.
Area of the circle is 113.1m.
Subtract A of circle from A of Square which equals 30.9meters. That is your answer.
If you want I can teach you step by step just comment if you want me to...
