6.1.8 The following set of scores was obtained from a quiz: 4, 5, 8, 9, 11, 13, 15, 18, 18, 18, 20. The teacher computes the usu
1 answer:
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Answer:
x = 3
y = 9
Explanation:
1- getting the value of x:
We are given that:
side AB is congruent to side DF. This means that:
AB = DF
3(2x+10) = 12x + 12
6x + 30 = 12x + 12
12x - 6x = 30 - 12
6x = 18
x = 18/6
x = 3
2- getting the value of y:
We are given that:
side BC is congruent to side FG. This means that:
BC = FG
2y + 12 = 2(2y-3)
2y + 12 = 4y - 6
4y - 2y = 12 + 6
2y = 18
y = 18/2
y = 9
Hope this helps :)
Answer:
x = 3
y = 9
Explanation:
1- getting the value of x:
We are given that:
side AB is congruent to side DF. This means that:
AB = DF
3(2x+10) = 12x + 12
6x + 30 = 12x + 12
12x - 6x = 30 - 12
6x = 18
x = 18/6
x = 3
2- getting the value of y:
We are given that:
side BC is congruent to side FG. This means that:
BC = FG
2y + 12 = 2(2y-3)
2y + 12 = 4y - 6
4y - 2y = 12 + 6
2y = 18
y = 18/2
y = 9
Hope this helps :)
Two main facts are needed here:
1. The logarithm , regardless of the base of the logarithm, exists for
.
2. The square root exists for
.
(in both cases we're assuming real-valued functions only)
By (2) we know that exists if
, or
.
By (1), we know that exists if
, or
. But as long as the square root exists, it will always be positive, so this condition will always be met.
Ultimately, then, we only require , so the function has domain
.
To determine the range, we need to know that, in their respective domains, and
increase monotonically without bound. We also know that
at minimum, at which point the square root term vanishes, so the least value the function takes on is
. Then its range would be
.