Answer:
I would change 2x+4y=24 into x=12–2y
To do that, divide both sides by 2 and then subtract 2y on each side.
After that, substitute for x. 3x+2y=19 would become 3(12–2y)+2y=19.
Then solve.
3(12–2y)+2y=19
36–6y+2y=19
-4y+36=19
-4y=-17
y=4.25
Then, substitute y in either equation.
Either this:
3x+2y=19
3x+2(4.25)=19
3x+8.5=19
3x=10.5
x=3.5
Or:
2x+4y=24
2x+4(4.25)=24
2x+17=24
2x=7
x=3.5
Or you could solve it in the equation you created in the beginning:
x=12–2y
x=12–2(4.25)
x=12–8.5
x=3.5
The coordinates where the lines intercept are (3.5, 4.25).
Sorry for the long answer!
Answer:
<em>The fraction of the beads that are red is</em>
Step-by-step explanation:
<u>Algebraic Expressions</u>
A bag contains red (r), yellow (y), and blue (b) beads. We are given the following ratios:
r:y = 2:3
y:b = 5:4
We are required to find r:s, where s is the total of beads in the bag, or
s = r + y + b
Thus, we need to calculate:
![\displaystyle \frac{r}{r+y+b} \qquad\qquad [1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Br%7D%7Br%2By%2Bb%7D%20%20%20%20%20%20%20%5Cqquad%5Cqquad%20%20%20%20%5B1%5D)
Knowing that:
![\displaystyle \frac{r}{y}=\frac{2}{3} \qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Br%7D%7By%7D%3D%5Cfrac%7B2%7D%7B3%7D%20%20%20%20%20%20%5Cqquad%5Cqquad%20%20%20%20%5B2%5D)
Multiplying the equations above:
Simplifying:
![\displaystyle \frac{r}{b}=\frac{5}{6} \qquad\qquad [3]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Br%7D%7Bb%7D%3D%5Cfrac%7B5%7D%7B6%7D%20%20%20%20%20%20%20%5Cqquad%5Cqquad%20%20%20%20%5B3%5D)
Dividing [1] by r:
Substituting from [2] and [3]:
Operating:
The fraction of the beads that are red is 
Answer:
i did a wild guess and i got the answer is 2
Step-by-step explanation:
2*7=14
14-4=10
10-2=8
Answer:
a₁ = 1
r = 3
Step-by-step explanation:
Since it’s a geometric series then
a₁ = 1 ( because 1 is the first term of the series)
3/1 = 9/3 = 27/9 = 81/27 = 3 then r=3.
Step-by-step explanation:
Yes it can be a direct variation. it follows the form y = kx
The only variable that cannot be held constant is F, but we can hold the mass constant and vary the acceleration OR we can hold the acceleration constant and vary the mass. Either one would work, but the easiest would be to vary the mass and hold the acceleration constant since we are all pulled by the same action gravity
