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3 years ago

A - g(x) = |x-5|+5 B - g(x) = |x-5| C - g(x) = |x|+5 D - g(x) = |x+5|+5

Mathematics

1 answer:

n200080 [17]3 years ago

5 0

Answer:

$g(x) = |x| + 5$

Step-by-step explanation:

Since both the original and transformed graph have the same x value for their vertex, there was no horizontal shift. However the vertex shifted up. Specifically it shifted up 5 units. Therefore:

$g(x) = f(x) + 5$

The plus 5 performs a translation up 5 units. Plugging in for f(x) it can be seen that:

$g(x) = |x| + 5$

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3 0

3 years ago

In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa

Elden [556K]

Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

$^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434$

Hence, the required number of ways =27434

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2 years ago