All categories

3 years ago

Find two ratios that are equivalent to 2/12

Mathematics

1 answer:

I am Lyosha [343]3 years ago

8 0

1/6
4/24

first one is half the denominator and numerator

second one is double

You might be interested in

Sasha has 4 quarters in her pocket. She has 6 times as many quarters in her piggy bank. Draw a picture to show the number of qua

WARRIOR [948]

Draw 24 quarters in her piggy bank

5 0

3 years ago

Read 2 more answers

Center city has a population of twenty one thousand seventy people write the populationin standard form

Digiron [165]

It would be 21070

In standard form, 20000+1000+70

5 0

3 years ago

Read 2 more answers

A total of 50 juniors and seniors were given a mathematics test. The 35 juniors attained an average score of 80 while the 15 sen

Alik [6]

Answer:

Step-by-step explanation:

Firstly, we need to calculate the total score of the junior students and the total score of the senior students.

The total score of the junior students is 35 * 80 = 2,800

The total score of the senior students is 15 * 70 = 1050

The total score is thus 2,800 + 1,050 = 3,850

The average score of the 50 students is thus 3,850/50 which equals 77

8 0

3 years ago

Helpppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp

BlackZzzverrR [31]

The answer is (-10/3)

3 0

3 years ago

Read 2 more answers

Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .

Solving for F(s) on the last equation, $F(s)=\frac{s}{s^2+25}$ , then the Laplace transform we were searching is $-F'(s)=\frac{s^2-25}{(s^2+25)^2}$

3 0

3 years ago