Exercise 1.2 1 Round each of these numbers to the degree of accuracy shown in brackets. a) 7765 (nearest hundred). b) 1099 (near
1 answer:
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Answer:
Explanation:
Since this is unfactorable, use the Quadratic Formula:
Now, take the information and input it:
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Answer:
Step-by-step explanation:
Given is a function as
Equating to 0 we have equation
If the function f(x) has x intercepts then the solutions are real
Let us use remainder theorem and change of signs rule
f(0) = 1>0
f(-1) = -1+3+1=3
f(-2) = -32+6+1<0
This implies there is a real root between -1 and -2.
f(1) = -1
Since f(0) and f(1) have different signs, there exists a real root between 0 and 1.
f(2) = 32-6+1>0
Since f(1) and f(2) have different signs there exists a real root between 1 and 2.
Thus there are definitely three real solutions as
one between -1 and 0, one between 0 and 1, and third between 1 and 2.