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3 years ago

Write an algebraic expression for the phrase the product of 4 and s.

Mathematics

2 answers:

notka56 [123]3 years ago

8 0

Answer: 4s

Step-by-step explanation:

Product means multiply

4 times s = 4s

padilas [110]3 years ago

5 0

Answer: 4 times s = 12

Step-by-step explanation:

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Estimate the limit. <br> Picture below

vova2212 [387]

Answer:

Hence, the limit of the expression: $\lim_{x \to 1} \dfrac{\dfrac{1}{x+2}-\dfrac{1}{3}}{x-1}$ is:

$\dfrac{-1}{9}$

Step-by-step explanation:

We are asked to estimate the limit of the expression:

$\lim_{x \to 1} \dfrac{\dfrac{1}{x+2}-\dfrac{1}{3}}{x-1}$

We will simplify the expression by first taking the l.c.m of the terms in the numerator to obtain the expression as:

$\dfrac{3-(x+2)}{3(x+2)}\\\\=\dfrac{3-x-2}{3(x+2)}\\\\=\dfrac{1-x}{3(x+2)}$

$\lim_{x \to 1} \dfrac{1-x}{3(x+2)(x-1)}\\\\= \lim_{x \to 1} \dfrac{-(x-1)}{3(x+2)(x-1)}\\\\\\= \lim_{x \to 1} \dfrac{-1}{3(x+2)}$

since the same term in the numerator and denominator are cancelled out.

Now the limit of the function exist as the denominator is not equal to zero when x→1.

Hence,

$\lim_{x \to 1} \dfrac{-1}{3(x+2)}\\\\=\dfrac{-1}{3(1+2)}\\\\=\dfrac{-1}{3\times 3}\\\\=\dfrac{-1}{9}$

Hence, the limit of the expression: $\lim_{x \to 1} \dfrac{\dfrac{1}{x+2}-\dfrac{1}{3}}{x-1}$ is:

$\dfrac{-1}{9}$

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