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3 years ago

Is zero an integer?please help me

Mathematics

2 answers:

julsineya [31]3 years ago

7 0

Yes it is all whole numbers are including negative numbers

dimaraw [331]3 years ago

7 0

Yes it is also a whole number

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a pair of jeans originally cost $140 and are now on sale for 65% off. what is the sale price of the jeans

Andrei [34K]

Answer:

$49

Step-by-step explanation:

Method 1:

100% - 65% = 35%

$\frac{y}{140}: \frac{35}{100}$

y · 100 = 35 · 140

100y = 4900

100y ÷ 100 = 4900 ÷ 100

y = 49

Method 2:

$\frac{y}{140}: \frac{65}{100}$

y · 100 = 65 · 140

100y = 9100

100y ÷ 100 = 9100 ÷ 100

y = 91

$140 - $91 = $49

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3 years ago

Find the 11th term 64,-32,16,-8

NeTakaya

Answer:

Step-by-step explanation:

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4 years ago

What type of errors do you think scientist make with measurements? Ready?

wariber [46]

Answer:

The errors that may occur in the measurement of a physical quantity can be classified into six types: constant error, systematic error, random error, absolute error, relative error and percentage error.

Step-by-step explanation:

Hope this will be helpful for you.

6 0

1 year ago

The distance between two points is 65‾√. One endpoint is (4,−3). It is known that the y-coordinate of the other endpoint is 7.

diamong [38]

Answer:

C & D

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Calculate the limit of the function with L'Hospital rule

mr_godi [17]

Answer:

L=24

Step-by-step explanation:

L'Hopital's Rule for Evaluating Limits:

Rule is that if $\lim_{x \to a} \frac{f(x)}{g(x)}$ takes $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, then,

$\lim_{x \to a} \frac{f(x)}{g(x)}= \lim_{x \to a} \frac{f'(x)}{g'(x)}$

where $f'(x)=\frac{df(x)}{dx}$ and $g'(x)=\frac{dg(x)}{dx}$

Now coming to the problem,

$L= \lim_{x \to \frac{\pi}{6} } \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3} )}$

Here $f(x)=cot^{3}x-3cotx$ and $g(x)=cos(x+\frac{\pi}{3} )$

Substituting $x=\frac{\pi}{6}$ in f(x) and g(x),

$f(\frac{\pi}{6})=cot^{3}\frac{\pi}{6}-3cot\frac{\pi}{6}\\=3\sqrt{3}-3\sqrt{3}\\ =0$

$g(\frac{\pi}{6})=cos(\frac{\pi}{6}+\frac{\pi}{3})\\=cos\frac{\pi}{2}\\=0$

Since L takes the form $\frac{0}{0}$ , using l'hopital's rule

$L= \lim_{x \to \frac{\pi}{6}} \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3})}= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}x(-cosec^{2}x)-3(-cosec^{2}x)}{-sin(x+\frac{\pi}{3})}$

now substituting $x=\frac{\pi}{6}$ ,

$L= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}\frac{\pi}{6}(-cosec^{2}\frac{\pi}{6})-3(-cosec^{2}\frac{\pi}{6})}{-sin(\frac{\pi}{6}+\frac{\pi}{3})}\\=\frac{3*3^{2}(-2^{2})+3(2^{2})}{-1}\\=24$

6 0

3 years ago