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3 years ago

The ratio of boys to girls in Ella's class is 2 to 5. There are 12 boys in Ella's class. How many girls are there? Answer value

Mathematics

2 answers:

sertanlavr [38]3 years ago

5 0

Answer:

There are 30 girls in Ella's class.

Step-by-step explanation:

Hope This Helps!

gizmo_the_mogwai [7]3 years ago

3 0

Answer:

x=30

Step-by-step explanation:

Given 2/5 , 12

2/5*12/x

Cross multiply

2x=60

x=30

Hope I helped!

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ASAP PLEASE HELP!!!!

zzz [600]

Answer:

It will take Josie 80 minutes to run 10 miles.

Step-by-step explanation:

We can set up a proportion between the number of minutes she runs to the number of miles she runs. From what we know in the equation, we know that it takes her 24 minutes to run 3 miles, and it takes x minutes to run 10 miles. We can represent this in a proportion!

We know that the ratio of 24:3 will be the same for any number of miles, so we can say

$\frac{24}{3} =\frac{x}{10\\}$

We can solve this by first simplifying the left hand side then isolation x by multiple both sides of the equation by 10.

24/3 = 8

8 = $\frac{x}{10}$

80 = x

It will take Josie 80 minutes to run 10 miles.

8 0

3 years ago

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Can u pls help me with this question

Shalnov [3]

Answer:

A, C, D

Step-by-step explanation:

7 0

3 years ago

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Every day Jack runs 3 km a) Find out how far he runs in 4 weeks

Dafna1 [17]

Answer:

84 km

Step-by-step explanation:

you times 7 by 4 and get 28 then you times 3 by 28 and get 84 km

7 0

2 years ago

Eric and Stephanie took their younger sister Melissa to pick apples. Eric picked 4 times as many apples as

vodka [1.7K]

Answer:

Melissa picked 16 apples

Step-by-step explanation:

(Eric and Stephanie) 160 divided by 10 (I put ten because I add 6 and 4)

Which will equal 16.

Hope this helps ~~~~ :)

4 0

3 years ago

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Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Ann [662]

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$ .

7 0

3 years ago