All categories

3 years ago

What are the LCM of 24 and 30?

Mathematics

2 answers:

Alla [95]3 years ago

8 0

The LCM of 24 and 30 is 120.

bogdanovich [222]3 years ago

3 0

Answer:

120

Step-by-step explanation:

You might be interested in

What is the slope of the line between (3, −4) and (−2, 1)? (1 point)

stealth61 [152]

-1 will be your answer

rise over run
3-1 = 2
-4 + 1 = -3
2-1 = 1
-3 + 1 = -2

hope this helps

8 0

3 years ago

DOES ANYONE KNOW HOW TO DO THIS???

Anna11 [10]

Answer:

Cost of a pound of chocolate chips: $3.5

Cost of a pound of walnuts: $1.25

Step-by-step explanation:

x - cost of a pound of chocolate chips

y - cost of a pound of walnuts

We create two equations based on the information we have:

3x+2y=13

8x+4y=33

The whole point of these problems os to get rid of x or y. In this question, we can do this by multiplying both sides of the first equation by 2, and then subtracting it from the second equation:

8x+4y=33

6x+4y=26

2x=7

x=3.5

Then we change x for 3.5 in the first equation:

3×3.5+2y=13

10.5+2y=13

2y=2.5

y=1.25

Hope this helps!

4 0

2 years ago

A polygon is shown:

Klio2033 [76]

Answer:

My brain just exploded. I tried imagining what the shape would look like, and that didn't work. Also, I tried drawing the shape and that didn't work. I have no idea what the shape looks like. If you still need help with this question, maybe you can repost it with an image attached so we can know what the shape looks like.

7 0

3 years ago

Read 2 more answers

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago

What is the average of 1/2 and 1/3

nadya68 [22]

Hi there!

In order to fin the average, you need to add all the terms together and then divide the result by the number of terms :

( $\frac{1}{2} + \frac{1}{3}$ ) ÷ 2 = average

Now, I’m assuming that you know that in order to add fractions, both fractions must have the same denominator (bottom number in a fraction). Since these fractions do not have the same denominator, we must give them one.

To find the lowest common denominator (which will help us solve this problem), we must find out what 2 & 3 go into. Well, both numbers go into 6!

So, if the denominator is now 6, you must multiply the numerator (numbers about the “/” line in this case are both 1) by how much you multiplied its denominator by.

For 1/2, you multiplied 2 by 3 to get 6. Therefore, you must multiply the 1 by 3 aswell.

for 1/3, you multiplied 3 by 2 to get 6. Therefore, you must multiply the 1 by 2.

and now you have 3/6+2/6 since the denominators are the same, you can add the numerators normally which gives you 5/6

$\frac{5}{6}$ ÷ 2 = $\frac{5}{12}$

Your answer is : $\frac{5}{12}$

There you go! I really hope this helped, if there's anything just let me know! :)

3 0

3 years ago