Answer:y=-4/5x+4
Step-by-step explanation:
I'm taking the liberty of editing your function <span>v = e5xey: It should be
</span>
<span>v = e^5x^ey, with " ^ " indicating exponentiation.
</span>
Did you mean e^(5x) or (e^5)x? I'll assume it's e^(5x).
The partial of v = e^(5x)e^y with respect to x is e^(5x)(5)*e^y, or 25x*e^y.
The partial of v = e^(5x)e^y with respect to y is e^(5x)e^y.
Answer:
x = 9.17 (nearest hundredth)
The variable x is the amount the student needs to save each month in addition to his usual saved amount of $20.
Step-by-step explanation:
350 = 12(x + 20)
Multiply out brackets: 350 = 12x + 240
Subtract 240 from both sides: 110 = 12x
Divide both sides by 12: 9 1/6 = x
x = 9.17 (nearest hundredth)
The variable x is the amount the student needs to save each month in addition to his usual saved amount of $20.
I think the answer for this question is the second one
The total amount of calories burnt in jogging 2 miles = 185 calories
We have to determine the number of calories burned in jogging 3 miles.
Firstly, we will determine the amount of calories burnt in jogging 1 mile. We will use unitary method to evaluate this.
So, the total amount of calories burnt in jogging 1 mile = 
= 92.5 calories
So, the amount of calories burnt in 3 miles = 
= 277.5 calories
Therefore, 277.5 calories are burned in jogging 3 miles.
