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AlexFokin [52]
2 years ago
11

Heyy i just need some help with questions 29 if anyone could help me and show the work that would be amazing thank you!!

Mathematics
1 answer:
horrorfan [7]2 years ago
3 0

Step-by-step explanation:

if we are taking in terms of domain.. is d set of number that we can put to the function with get division by zero .. squaroot of negative number,log of zero e.t.c

A) p(x)/m(x)= 1/squaroot(x) ÷ x²-4 which we know that x²-4 must not be equal to zero

because if is equal to zero we will have division by zero, which have contradicted the hypothesis of law of domain

x²-4 =! 0

(x-2)(x+2)=!0

x=!2 or x=!-2 because if is equal to that it's has contradicted d hypothesis

the answer is all real number except -2 and -

2

B)p(m(x))=p(x²-4) =1/squaroot of x²-4

note they are two things involve we must not get division by zero and root of negative number

we have all real number expect from -2 to 2

C)m(p(x))=m(1/root of (x))=(1/root of (x))²_4)

we have 1/x-4

so the answer is all real number expect zero

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Step-by-step explanation:

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Which is equal to P-R?
lorasvet [3.4K]
<h3>Answer: Choice B</h3>

With matrix subtraction, you simply subtract the corresponding values.

I like to think of it as if you had 2 buses. Each bus is a rectangle array of seats. Each seat would be a box where there's a number inside. Each seat is also labeled in a way so you can find it very quickly (eg: "seat C1" for row C, 1st seat on the very left). The rule is that you can only subtract values that are in the same seat between the two buses.

So in this case, we subtract the first upper left corner values 14 and 15 to get 14-15 = -1. The only answer that has this is choice B. So we can stop here if needed.

If we kept going then the other values would be...

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The values in bold correspond to the proper values shown in choice B.

As you can probably guess by now, matrix addition and subtraction is only possible if the two matrices are the same size (same number of rows, same number of columns). The matrices don't have to be square.

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