Answer:
(x-4) is not a factor of f(x)=x³-2x²+5x+1
Step-by-step explanation:
Uding the remainder theorem,(x-4) is a factor if the remainder is 0
Plug in x=4
(4)³-2(4)²+5(4)+1
64-32+20+1
64-53
11
A walking stick is way longer than a bumble bee
0.77875...
which, I'm assuming, they want you to round to the nearest hundredth? So, .78?...
We have to present the number 41 as the sum of two squares of consecutive positive integers.
1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
16 + 25 = 41
<h3>Answer: 4 and 5</h3>
Other method:
n, n + 1 - two consecutive positive integers
The equation:
n² + (n + 1)² = 41 <em>use (a + b)² = a² + 2ab + b²</em>
n² + n² + 2(n)(1) + 1² = 41
2n² + 2n + 1 = 41 <em>subtract 41 from both sides</em>
2n² + 2n - 40 = 0 <em>divide both sides by 2</em>
n² + n - 20 = 0
n² + 5n - 4n - 20= 0
n(n + 5) - 4(n + 5) = 0
(n + 5)(n - 4) = 0 ↔ n + 5 = 0 ∨ n - 4 =0
n = -5 < 0 ∨ n = 4 >0
n = 4
n + 1 = 4 + 1 = 5
<h3>Answer: 4 and 5.</h3>
