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3 years ago

7

Gabriel was out at a restaurant for dinner when the bill came. His dinner came to $20. After adding in a tip, before tax, he pai

d $23. Find the percent tip.

1 answer:

swat323 years ago

4 0

Answer:

Step-by-step explanation: I think it is three percent

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50 - [ 7 + ( 3^(2) / 2 ) ]<br> Step by step please

Butoxors [25]

Answer:

evaluation : 38.5 38.5 is the answer because you simplify this term

Step-by-step explanation:

8 0

3 years ago

\lim _{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)

Vinvika [58]

$\displaystyle \lim_{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)$

Both the numerator and denominator approach 0, so this is a candidate for applying L'Hopital's rule. Doing so gives

$\displaystyle \lim_{x\to 0}\left(2\ln(1+3x)+\dfrac{6x}{1+3x}+\cos(x)\tan(3x)+3\sin(x)\sec^2(x)-6x^2}{3\sin(3x)}\right)$

This again gives an indeterminate form 0/0, but no need to use L'Hopital's rule again just yet. Split up the limit as

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} + \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} \\\\ + \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} + \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} \\\\ - \lim_{x\to0}\frac{6x^2}{3\sin(3x)}$

Now recall two well-known limits:

$\displaystyle \lim_{x\to0}\frac{\sin(ax)}{ax}=1\text{ if }a\neq0 \\\\ \lim_{x\to0}\frac{\ln(1+ax)}{ax}=1\text{ if }a\neq0$

Compute each remaining limit:

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{\ln(1+3x)}{3x} \times \lim_{x\to0}\frac{3x}{\sin(3x)} = \frac23$

$\displaystyle \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\frac{1}{1+3x} = \frac23$

$\displaystyle \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\cos(x)}{\cos(3x)} = \frac13$

$\displaystyle \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\sin(x)}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\sec^2(x) = \frac13$

$\displaystyle \lim_{x\to0}\frac{6x^2}{3\sin(3x)} = \frac23 \times \lim_{x\to0}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}x = 0$

So, the original limit has a value of

2/3 + 2/3 + 1/3 + 1/3 - 0 = 2

6 0

3 years ago

What is the range of the function on the graph

Tanzania [10]

Answer:

All real numbers greater than or equal to -2.

Step-by-step explanation:

The range is the set of values of the y-coordinates of the points on the graph.

The smallest y-coordinate is -2. All real numbers greater than equal to -2 are the range.

Answer: All real numbers greater than or equal to -2.

6 0

3 years ago

W+(−2.8)=4.3 whats the answer

IgorLugansk [536]

Answer:

w=7.1

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Which of the following represents the characteristics of the graph of x ≥ -9 1.) closed circle on -9 and shaded to the left 2.)

Harrizon [31]

Answer:

Closed circle on -9 and shaded to he right

8 0

3 years ago