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Neko [114]
3 years ago
15

I don't know how to do it

Mathematics
2 answers:
kow [346]3 years ago
7 0
Alright. So the question basically states everything for you. p is less than q. So just make up numbers that satisfy this. Lets say p=3 and q=7. And r=2. Just plug in the numbers now. p+r=5 and q+r=9. 
A) says r-q is how much more less p+r is than q+r. When you do r-q, you get -5. But the difference between p+r and q+r isn't -5.
B) says that q+p is how much more less p+r is than q+r. But when you do q+p, you get 10, which isn't the difference.
C) says q-p is how much more less p+r is than q+r. When you plug in the numbers, for q-p, you get 4, which makes sense (correct answer).
Now, since you already know the answer's C, there's no real point in trying out answer choice D. Hope this helped! 
kolezko [41]3 years ago
4 0
Please elaborate more on your question so I can help you
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steposvetlana [31]

Answer:

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Step-by-step explanation:

The limit is:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}

by replacing, and applying repeatedly you obtain:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}

hence, the limit of the function is -1/14

8 0
3 years ago
What is the equation of the line that passes through (1, 2) and is parallel to the line whose equation is 4x + y +1 = 0?
tamaranim1 [39]
6/9 cs it the right one
6 0
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Given h(x) - 2x + 9, find the value of x when h(x) - 21.<br> A) 51<br> B) 12<br> C) 6<br> D) 30
Akimi4 [234]

Answer:

C, 6!

Step-by-step explanation:

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4 0
3 years ago
Can someone help me pleaseeeeeeeeee ASAP
lord [1]

Answer:

18/23

Step-by-step explanation:

36/46

then simplified to 18/23

5 0
3 years ago
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3 years ago
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