All categories

3 years ago

Select the fraction that is equal to 2/4 1/5 12/7 1/2

Mathematics

1 answer:

polet [3.4K]3 years ago

6 0

Answer:

1/2

hope this helps

have a good day :)

Step-by-step explanation:

You might be interested in

A number divided by and then the quotient is increased by 8. the result is 33. write and solve an equation to find each number

yanalaym [24]

Answer:

do 8 times 4.2 then subtract the .6 to get to 33 hope this helped

5 0

3 years ago

Help<br> 1<br> A /<br> 7 (x + 2) -5 = 30

DaniilM [7]

Answer:

x = 3

Step-by-step explanation:

Given

7(x + 2) - 5 = 30 ( add 5 to both sides )

7(x + 2) = 35 ( divide both sides by 7 )

x + 2 = 5 ( subtract 2 from both sides )

x = 3

3 0

3 years ago

These angels are equal. What are they called?

Yanka [14]

Acute angles because they are less then 90 degrees which is a right angle.

7 0

3 years ago

Need help a s a p HURRYYYYY

Dimas [21]

Step-by-step explanation:

v= 4/3πr³

4/3×314/100×8×8×8

•

= 2,143.573

3 0

3 years ago

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

1 year ago