<h3>Answer: Choice C. </h3><h3>Division[ (4x^3+2x^2+3x+5)^2, x^2+3x+1]</h3>
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Explanation:
It honestly depends on the CAS program, but for GeoGebra for instance, the general format would be Division[P, Q]
Where,
- P = numerator = (4x^3+2x^2+3x+5)^2
- Q = denominator = x^2+3x+1
As another example, let's say we want to divide x^2+5x+6 all over x^3+7 as one big fraction
We would type in Division[x^2+5x+6, x^3+7]
Answer:
x = 2 and y = 1
Step-by-step explanation:
A) 5x+y=11
B) x - y = 1.
(Solving by substitution)
B) x - 1 = y
A) 5x + (x -1) = 11. 6x-1=11. 6x = 11+1
6x = 12
x = 12/2. x = 2
B. 2 - 1 = y. y = 1
An injective function is not a function that is surjective. This means that you want a function that has a unique output for each input, that doesn't cover the natural numbers.
In formal terms a function [Math Processing Error] is injective if [Math Processing Error] implies [Math Processing Error].
We also know that it's not surjective because no value maps to [Math Processing Error] (or any odd number) since if [Math Processing Error], then [Math Processing Error]. However, since [Math Processing Error], the function isn't surjective.
Answer is B.
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Answer: First one: lies in the shaded region of the top inequality, but not in the shaded region of the bottom inequality.
Y=mx+b
slope=m
yint=b
we are given
slope=-2/3
point=(-3,-1)
y=-2/3x+b
input point -3,-
x=-3
y=-1
find b
-1=-2/3(-3)+b
-1=2+b
minus 3
-3=b
y=-2/3x-3
answer is D