You can create the equation 1.20x + 2.50y = 300
Where 1.20 is the price of each vinca flower, x is how many vinca you will buy, 2.50 is the price of each phlox flower, y is how many plhox you will buy and 300 is how much money you have.
To get three combinations of how many plhox and vinca flowers you ( or the gardner can buy) with 300, you can just take a random value for x or y, apply it to the equation and then solve the equation for the other
variable ( x or y).
This means:
Pick a number of how many flower you will get of ONE type of flower ( ONLY ONE)
The gardner will not get any Vinca flowers, then x = 0
( as x = how many vinca flower the gardner will get)
now apply 0 to x ( this means substitute x with 0)
(1.20)(0) + 2.50y = 300
Since any number x 0 is 0, the equation will be
0 + 2.50y = 300 or just 2.50y = 300
Now to solve an equation we just need to isolate the variable (y) on one side of the equation and the value on the other.
We can do that by dividing both sides of the equation, so 2.50y turns into 2.50/2.50 times y which is just y.
2.50/2.50 y = 300/2.50
y = 120
And since we chose the value for how many vincas we would buy, we have the x value as well ( 0)
x = 0
So 1 combination can be: 0 Vinca and 120 Plhox
Combination #2 ( I ll stop explaining the steps since I already explained how to do it)
y = 0 ( I chose not to buy any Plhox)
1.20x + (2.50)(0) = 300
1.20x = 300
1.20/120x = 300/120x
x = 25
y= 0
Combination 2: 25 Vincas and 0 plhox
Combination #3:
x = 100 ( I chose to buy 10 vincas)
(1.20)(100) + 2.50y = 300
120 + 2.50y = 300
( I didnt explain thow to isolate a variable in this situation so ill explain it: to isolate this esquation we can subtract 12 from both sides of the equation so 12 cancels out. We need to do it on both sides because otherwise the first side wouldnt be equal to the other sode, and that wouldnt be an equation)
120 - 120 + 2.50y = 300 - 120
2.50y = 180
y = 72
x = 100
Combination 3: 100 Vinca and 72 Plox
I know this is long but I hope it was helpful to you if you didnt know how to solve equations :)
Answer:
a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4
Step-by-step explanation:
for the equation
(1 + x⁴) dy + x*(1 + 4y²) dx = 0
(1 + x⁴) dy = - x*(1 + 4y²) dx
[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx
∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx
now to solve each integral
I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁
I₂= ∫[-x/(1 + x⁴)] dx
for u= x² → du=x*dx
I₂= ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ = - tan⁻¹ (x²) +C₂
then
1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) +C
for y(x=1) = 0
1/2 *tan⁻¹ (2*0) = - tan⁻¹ (1²) +C
since tan⁻¹ (1²) for π/4+ π*N and tan⁻¹ (0) for π*N , we will choose for simplicity N=0 . hen an explicit solution would be
1/2 * 0 = - π/4 + C
C= π/4
therefore
1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4
Divide the numerator and denominator by their gcf (25)
75/25= 3
100/25= 4
Final answer: 3/4
Step-by-step explanation:
I hope this is your answer
