n = {2, 5, 0} holds true for 2n - 3 < 9
<em><u>Solution:</u></em>
Replacement set is the set of values that may be substituted for the variable.
Plug in each value from the replacement set and evaluate both sides of the inequality
If the inequality is true for a certain value, that value belongs in the solution set.
<em><u>Given equation is:</u></em>
2n - 3 < 9
<em><u>Substitute n = 2</u></em>
2(2) - 3 < 9
4 - 3 < 9
1 < 9
1 is less than 9 is true
Thus n = 2 is a solution to equation
<em><u>Substitute n = 5</u></em>
2(5) - 3 < 9
10 - 3 < 9
7 < 9
7 is less than 9 is true
Thus n = 5 is a solution to equation
<em><u>Substitute n = 0</u></em>
2(0) - 3 < 9
0 - 3 < 9
-3 < 9
-3 is less than 9 is true
Thus n = -3 is a solution to equation
We are giiven with two realms: 36 students who passed test 1 and <span>38 who passed test 2 in which 42 students overall took the test. to determine U' or those who didn't pass either of the tests, then we use the formula:
A or B = (A+B)-AandB
A/B = (36+38)-AandB= 74-AandB
42 = A+ B - AandB + U'; 42 = 74- AandB+ U'
where U' is the number who did not pass
U'= -32 + AandB
U' cannot be negative but can be zero,
AandB = 32 @ minimum number of those who fail.
AandB cannot be greater than 38 but can be greater than 36, hence
U' = -32 + 38 @ maximum number of those who fail.
Answer then is 6.
</span>
I think it's C but don't quote me on that I don't know
9514 1404 393
Answer:
1) f⁻¹(x) = 6 ± 2√(x -1)
3) y = (x +4)² -2
5) y = (x -4)³ -4
Step-by-step explanation:
In general, swap x and y, then solve for y. Quadratics, as in the first problem, do not have an inverse function: the inverse relation is double-valued, unless the domain is restricted. Here, we're just going to consider these to be "solve for ..." problems, without too much concern for domain or range.
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1) x = f(y)
x = (1/4)(y -6)² +1
4(x -1) = (y-6)² . . . . . . subtract 1, multiply by 4
±2√(x -1) = y -6 . . . . square root
y = 6 ± 2√(x -1) . . . . inverse relation
f⁻¹(x) = 6 ± 2√(x -1) . . . . in functional form
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3) x = √(y +2) -4
x +4 = √(y +2) . . . . add 4
(x +4)² = y +2 . . . . square both sides
y = (x +4)² -2 . . . . . subtract 2
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5) x = ∛(y +4) +4
x -4 = ∛(y +4) . . . . . subtract 4
(x -4)³ = y +4 . . . . . cube both sides
y = (x -4)³ -4 . . . . . . subtract 4
It helps you visualization