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.
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-12.6 as a percent is -1260%
a)
Answer: 0.91 m
Explanation:
We know that,
P.E. = m g h
Where,
P.E = Potential energy
m = Mass of the object
g = acceleration due to gravity (9.8 m/s²)
It is given that, m = 1.5 kg
P.E. = 13.44 J
⇒ 13.44 = 1.5 kg × 9.8 m/s² × h
⇒ h = 0.91 m
Hence, apple sits om 0.91 m tall counter.
b)
Answer: 216 J
Explanation:
P.E. = m g h
Weight, mg = 120 N ( given)
height, h = 1.8 m ( given)
The energy possessed by the suitcase is due to virtue of its position (gravitational potential energy)
P.E. = 120 N × 1.8 m = 216 J
Hence, the energy possessed by the suitcase sitting on the counter is 216 J.
The angle that is same side exterior angle to ∠ABC is ∠EFH.
<h3>How to find angles?</h3>
When parallel line are cut by a transversal line, angle relationships are formed such as corresponding angles, vertically opposite angles, alternate angles, same interior angles, etc.
AD and EG are parallel to each other.
CH is a transversal line to the parallel lines.
Therefore, let's find the angle that has the same relationship as same side exterior angles as angle ABC.
Two angles that are exterior to the parallel lines and on the same side of the transversal line are called same-side exterior angles.
Same side exterior angles are supplementary.
Therefore,
∠ABC and ∠EFH are same side exterior angles.
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Let "a" and "s" represent the costs of advance and same-day tickets, respectively. Your problem statement gives you two relations.
.. a + s = 35 . . . . . the combined cost of one of each is 35
.. 15a +40s = 900 . . total paid for this combination of tickets was 900
There are many ways to solve these equations. You've probably been introduced to "substitution" and "elimination" (or "addition"). Using substitution for "a", we have
.. a = 35 -s
.. 15(35 -s) +40s = 900 . . substitute for "a"
.. 25s +525 = 900 . . . . . . . simplify
.. 25s = 375 . . . . . . . . . . . .subtract 525
.. s = 15 . . . . . . . . . . . . . . .divide by 25
Then
.. a = 35 -15 = 20
The price of an advance ticket was 20.
The price of a same-day ticket was 15.
