Answer:
Yes
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Coordinates (x, y)
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>
Point (7, 0)
Equation y = x - 7
<u>Step 2: Find</u>
- Substitute in point [Equation]: 0 = 7 - 7
- Subtract: 0 = 0
Since 0 = 0 is true, then it would be a solution to the equation.
The first solution is quadratic, so its derivative y' on the left side is linear. But the right side would be a polynomial of degree greater than 1, so this is not the correct choice.
The third solution has a similar issue. The derivative of √(x² + 1) will be another expression involving √(x² + 1) on the left side, yet on the right we have y² = x² + 1, so that the entire right side is a polynomial. But polynomials are free of rational powers, so this solution can't work.
This leaves us with the second choice. Recall that
1 + tan²(x) = sec²(x)
and the derivative of tangent,
(tan(x))' = sec²(x)
Also notice that the ODE contains 1 + y². Now, if y = tan(x³/3 + 2), then
y' = sec²(x³/3 + 2) • x²
and substituting y and y' into the ODE gives
sec²(x³/3 + 2) • x² = x² (1 + tan²(x³/3 + 2))
x² sec²(x³/3 + 2) = x² sec²(x³/3 + 2)
which is an identity.
So the solution is y = tan(x³/3 + 2).
-468
Explanation: Substitute in the numbers for the variables presented. I’ve shown it step-by-step in the photo below.
He answer of this question is 17
If you have these conditions:
1. (r ,theta) where r > 0
<span>2. (r, theta) where r < 0
</span>
The solution would be:
<span>r = sqrt(x^2 + y^2)
t = arctan(y/x)
r = sqrt(12 + 4) = sqrt(16) = +/- 4
t = arctan(2 / -2sqrt(3)) = arctan(-1 / sqrt(3)) = 5pi/6 , 11pi/6
1)
r > 0
(4 , 11pi/6)
2)
r < 0
(-4 , 5pi/6)
</span>
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