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Yakvenalex [24]
3 years ago
10

For what value(s) of k will the relation not be a function

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
5 0

We are given two relations

(a)

Relation (R)

R=[((k-8.3+2.4k),-5),(-\frac{3}{4}k,4)]

We know that

any relation can not be function when their inputs are same

so, we can set both x-values equal

and then we can solve for k

k-8.3+2.4k=-\frac{3}{4} k

3.4k-8.3=-\frac{3}{4}k

3.4k\cdot \:10-8.3\cdot \:10=-\frac{3}{4}k\cdot \:10

4k-83=-\frac{15}{2}k

34k=-\frac{15}{2}k+83

\frac{83}{2}k=83

83k=166

k=2............Answer

(b)

S = {(2−|k+1| , 4), (−6, 7)}

We know that

any relation can not be function when their inputs are same

so, we can set both x-values equal

and then we can solve for k

2-|k+1|=-6

2-\left|k+1\right|-2=-6-2

-\left|k+1\right|=-8

\left|k+1\right|=8

Since, this is absolute function

so, we can break it into two parts

|f\left(k\right)|=a\quad \Rightarrow \:f\left(k\right)=-a\quad \mathrm{or}\quad \:f\left(k\right)=a

k+1=-8\quad \quad \mathrm{or}\quad \:\quad \:k+1=8

we get

k+1=-8\quad

k=-9

k+1=8\quad

k=7

so,

k=-9\quad \mathrm{or}\quad \:k=7...............Answer

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z = - 0.8

Using the standard distribution table for z:

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2nd: Calculating for z when x = 14.6 lb

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z = 1.2

Using the standard distribution table for z:

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