Help me pls

Mathematics

1 answer:

expeople1 [14]3 years ago

5 0

I think you can find it graphically by points through a table you create like:

let x = 0, and find the value of y

for example for the first equation when we let x = 0 you find y = 6 and be in the form of (0,6) then draw them or you can let x = -2 or -3 , etc as you like

to be more accurate you can find 3 pionts and start graphing them

these two equations are straight lines and they will intersect in a point

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Use the given facts about the functions to find the indicated limit.

oksian1 [2.3K]

Answer: Option b.

Step-by-step explanation:

1. You can rewrite it as following:

$\lim_{x\to \ -6} 2$

This means that you must substitute x=-6 into the limit. There is not x, then the function is always equal to 2, no matter the value of x.

$\lim_{x\to \ -6} 2=2$

$\lim_{x\to \ -6} 1$

This means that you must substitute x=-6 into the limit. There is not x, then the function is always equal to 1, no matter the value of x.

$\lim_{x\to \ -6} 1=1$

2. Then:

$\lim_{x\to \ -6} (f/g)(x)=2/1=2$

3. The answer 2.

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3 years ago

Read 2 more answers

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 :-

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