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3 years ago

Evaluate 1/3 ÷ 2/5 = Give your answer as a fraction in its simplest form.

Mathematics

2 answers:

Softa [21]3 years ago

3 0

Answer:

5/6

Step-by-step explanation:

Since 1/3÷2/5 can also be written as 1/3/2/5

So the ÷ can then be converted to × and 2/5 will change to 5/2

So let's solve

1/3×5/2

5/6

Since there is no common number that can divide the two

So the final answer is 5/6

salantis [7]3 years ago

3 0

Answer:

5/6

Step-by-step explanation:

1/3 divided by 2/5 is 2 1/2

this is because when you do this you use KCF

Keep Change Flip

you keep 1/3

change the sign to times

and flip the last fraction around to 5/2

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5 0

3 years ago

Peter is x years old and his sister is 5 years older. If the product of their ages is 84, form an equation in x and solve it to

daser333 [38]

Answer:

x^2 + 5x - 84 = 0

x = 7

Step-by-step explanation:

let the sister's age = y

sister's age (y) = 5 + x equation 1

xy = 84 equation 2

Make y the subject of the formula in equation 2

y = 84/x

equate equation 1 and 2

84/x = 5 + x equation 3

Multiply equation 3 by x

84 = 5x + x^2

x^2 + 5x - 84 = 0

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x^2 -7x + 12x -84 = 0

x(x + 12) -7(x + 12)

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3 0

3 years ago

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