Answer:
d³y/dx³ = (-2xy² − 3x³ − 4xy²) / (8y⁵)
Step-by-step explanation:
d²y/dx² = (-2y² − x²) / (4y³)
Take the derivative (use quotient rule and chain rule):
d³y/dx³ = [ (4y³) (-4y dy/dx − 2x) − (-2y² − x²) (12y² dy/dx) ] / (4y³)²
d³y/dx³ = [ (-16y⁴ dy/dx − 8xy³ − (-24y⁴ dy/dx − 12x²y² dy/dx) ] / (16y⁶)
d³y/dx³ = (-16y⁴ dy/dx − 8xy³ + 24y⁴ dy/dx + 12x²y² dy/dx) / (16y⁶)
d³y/dx³ = ((8y⁴ + 12x²y²) dy/dx − 8xy³) / (16y⁶)
d³y/dx³ = ((2y² + 3x²) dy/dx − 2xy) / (4y⁴)
Substitute:
d³y/dx³ = ((2y² + 3x²) (-x / (2y)) − 2xy) / (4y⁴)
d³y/dx³ = ((2y² + 3x²) (-x) − 4xy²) / (8y⁵)
d³y/dx³ = (-2xy² − 3x³ − 4xy²) / (8y⁵)
B) the values will create a line with positive slope when graphed on a scatter plot.
There are 600 students including the seventh and eighth graders at the party.
This problem uses the concept of percentages to define the conditions that are laid in front of us.
Let the original number of students be S , and the number of seventh graders be = 0.60S
We know that percent is used to convey the mathematical term of a fraction multiplied by 100.
Total students after 20 eighth graders arrive = S + 20
And we have that
Number of seventh graders / total number of students = 58%
.60S / [ S + 20 ] = .58 we multiply both sides by S + 20
0.60S =0 .58 [ S + 20]
.60S = .58S + 11.6 we subtract 0.58S from both the sides
0.02S = 11.6 we divide both the sides by .02
S = 11/6 / 0.02 = 580
So the total number of students = 580 + 20 = 600 .
Hence there are 600 students at the party at that time.
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Answer:
Step-by-step explanation:
-x-12
The answer to this equation is 45
