What do you need help on?
Answer: Quest 1 final result. (I looked all this up)
(2x -1) * (5x squared - 2x + 6)
_____________________.
4
I could not find the work for part two but I got the answer: x=3,7
Answer:
There are 10 counters in the bag and you want the 4
As their is only one for the probability is
1/10
There are now 9 counters left in the bag.
You only want the even ones , which are 2 6 8 and 10 ( four has been taken out)
There are only four even counters in the bag of nine counters so the probability is
4/9
As you want both of these to occur, you need to multiply them
1/10 x 4/9 = 49/90
Step-by-step explanation:
1 ) 9 x + 18 > 9 x - 27
9 x - 9 x > - 27 - 18
0 * x > - 45
Always true
2 ) 6 x - 13 < 9 x - 12
6 x - 6 x < - 12 + 13
0 * x < 1
Always true
3 ) - 6 ( 2 x - 10 ) + 12 x ≤ 180
- 12 x + 60 + 12 x ≤ 180
0 * x ≤ 180 - 60
0 * x ≤ 120
Always true.
The smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
What is the intermediate value theorem?
Intermediate value theorem is theorem about all possible y-value in between two known y-value.
x-intercepts
-x^2 + x + 2 = 0
x^2 - x - 2 = 0
(x + 1)(x - 2) = 0
x = -1, x = 2
y intercepts
f(0) = -x^2 + x + 2
f(0) = -0^2 + 0 + 2
f(0) = 2
(Graph attached)
From the graph we know the smallest positive integer value that the intermediate value theorem guarantees a zero exists between 0 and a is 3
For proof, the zero exists when x = 2 and f(3) = -4 < 0 and f(0) = 2 > 0.
<em>Your question is not complete, but most probably your full questions was</em>
<em>Given the polynomial f(x)=− x 2 +x+2 , what is the smallest positive integer a such that the Intermediate Value Theorem guarantees a zero exists between 0 and a ?</em>
Thus, the smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
Learn more about intermediate value theorem here:
brainly.com/question/28048895
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