I'm assuming you have to graph a line that shows the combination of earnings.
Our equation is 14.64y + 12.20x = 219.60
We'll be solving for x and y. We'll need to substitute the variables, one at a time, to 0, in order to find their intercepts.
Let's do x first.
14.64(0) + 12.20x = 219.60
12.20x = 219.60
Divide by 12.20 on both sides.
x = 18
Now let's solve for y.
14.46y + 12.20(0) = 219.60
14.46y = 219.60
Divide both sides by 14.46.
y = 15.18
That is actually a long decmal, but since we're dealing with money, we round it down to the nearest cent.
So, now we know x = 18 and y = 15.18.
Graph those on the x coordinate plane and the y coordinate plane, and make a line connecting the two. That line represents all the solutions.
Answer: just remove the decimal so 350/ 76 =
Step-by-step explanation: easy
This is a little long, but it gets you there.
- ΔEBH ≅ ΔEBC . . . . HA theorem
- EH ≅ EC . . . . . . . . . CPCTC
- ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
- ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
- ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
- ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
- ΔDAC ≅ ΔDAG . . . HA theorem
- DC ≅ DG . . . . . . . . . CPCTC
- ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
- ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
- ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
- ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
- (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
- ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
- This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
Answer:
55860
Step-by-step explanation:
This way super dupper easy
-5-(15y-1) = 2(7y-16) - y // question
-5-15y+1 = 14y-32-y // distribute
-15y - 4 = 13y - 32 // subtract
-28y = -28 // divide
y = 1
