I cant do this anymore

3x-2y=4<br><br> How do you do this using common core ?

bixtya [17]

Graph 3x+2y=4.
3x+2y=4 3<span> x + 2 y = 4.</span>
Solve for y y .
<span>Since 3x 3 ⁢ x does not contain the variable to solve for, move it to the right side of the equation by subtracting 3x 3 ⁢ x from both sides.</span>

3 0

3 years ago

Read 2 more answers

Your little cousin will have a big animal-themed birthday party and you will help with the goody bags. Each bag will have a penc

Bas_tet [7]

Using the least common factor, it is found that:

a) 60 packages should be bought.

b) There will be 5 filled goody bags.

<h3>Least Common Factor:</h3>

The sizes of the packages are: 10, 6, 15 and 12.

To fill each bag and have no left-overs, the number of packages is the <u>least common factor</u> of these amounts.

The least common factor is found factoring the numbers into prime factors.

Item a:

10 - 6 - 15 - 12|2

5 - 3 - 15 - 6|2

5 - 3 - 15 - 3|3

5 - 1 - 5 - 1|5

1 - 1 - 1 - 1

Hence, lcf(10,6,15,2) = 2 x 2 x 3 x 5 = 60.

60 packages should be bought.

Item b:

Goody bags are in packages of 12, hence:

60/12 = 5.

There will be 5 filled goody bags.

To learn more about the least common factor, you can take a look at brainly.com/question/24873870

6 0

2 years ago

A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.50 kg mass in each outstre

Vlad1618 [11]

Answer:r'=0.327 m

Step-by-step explanation:

Given

$N=2.85 rev/s$

angular velocity $\omega =2\pi N=17.90 rad/s$

mass of objects $m=1.5 kg$

distance of objects from stool $r_1=0.789 m$

Combined moment of inertia of stool and student $=5.53 kg.m^2$

Now student pull off his hands so as to increase its speed to 3.60 rev/s

$\omega _2=2\pi N_2$

$\omega _2=2\pi 3.6=22.62 rad/s$

Initial moment of inertia of two masses $I_0=2mr_^2$

$I_0=2\times 1.5\times (0.789)^2=1.867$

After Pulling off hands so that r' is the distance of masses from stool

$I_0'=2\times 1.5\times (r')^2$

Conserving angular momentum

$I_1\omega =I_2\omega _2$

$(5.53+1.867)\cdot 17.90=(5.53+I_o')\cdot 22.62$

$I_0'=1.397\times 0.791$

$I_0'=5.851$

$5.53+2\times 1.5\times (r')^2=5.851$

$2\times 1.5\times (r')^2=0.321$

$r'^2=0.107009$

$r'=0.327 m$

7 0

3 years ago

What should I say??!?!?!?!?!?!??!

djyliett [7]

Answer:

The North and South poles on the bar magnets attract to one another when they are placed near each other. When the North and North or the south and south are placed near each other, they repel away from each other.

I hope this helps :)

8 0

3 years ago

I dont understand the problem

Roman55 [17]

Answer:

equation of a line:

y = mx+c

1) find the gradient, m

$m = \frac{y2 - y1}{x2 - x1}$

$m = \frac{ - 4 - ( - 4)}{ - 10 - 1}$

$m = 0$

2) find y-intercept, c using coordinate (1,-4)

y = mx + c

-4 = 0(1) + c

c = -4

the equation of line:

y = mx+c

y = 0(x) + c

y = c

y = -4

6 0

3 years ago