Answer:
y - 5 = -2(x + 3)
Step-by-step explanation:
When you write an equation in point-slope form, you only need two things: a point and a slope.
Given:
point: (-3, 5)
slope (m): -2
The standard point-slope equation is
y - y₁ = m(x - x₁)
Plug in what you know.
y - (5) = -2(x - (-3))
Simplify.
y - 5 = -2(x + 3)
This is your equation.
Learn with another example:
brainly.com/question/24436844
The Jacobian for this transformation is
![J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}](https://tex.z-dn.net/?f=J%20%3D%20%5Cbegin%7Bbmatrix%7D%20x_u%20%26%20x_v%20%5C%5C%20y_u%20%26%20y_v%20%5Cend%7Bbmatrix%7D%20%3D%20%5Cbegin%7Bbmatrix%7D%204%20%26%200%20%5C%5C%200%20%26%203%20%5Cend%7Bbmatrix%7D)
with determinant
, hence the area element becomes
![dA = dx\,dy = 12 \, du\,dv](https://tex.z-dn.net/?f=dA%20%3D%20dx%5C%2Cdy%20%3D%2012%20%5C%2C%20du%5C%2Cdv)
Then the integral becomes
![\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_%7BR%27%7D%204x%5E2%20%5C%2C%20dA%20%3D%20768%20%5Ciint_R%20u%5E2%20%5C%2C%20du%20%5C%2C%20dv)
where
is the unit circle,
![\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E2%7D%7B16%7D%20%2B%20%5Cdfrac%7By%5E2%7D9%20%3D%20%5Cdfrac%7B%284u%5E2%29%7D%7B16%7D%20%2B%20%5Cdfrac%7B%283v%29%5E2%7D9%20%3D%20u%5E2%20%2B%20v%5E2%20%3D%201)
so that
![\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv](https://tex.z-dn.net/?f=%5Cdisplaystyle%20768%20%5Ciint_R%20u%5E2%20%5C%2C%20du%20%5C%2C%20dv%20%3D%20768%20%5Cint_%7B-1%7D%5E1%20%5Cint_%7B-%5Csqrt%7B1-v%5E2%7D%7D%5E%7B%5Csqrt%7B1-v%5E2%7D%7D%20u%5E2%20%5C%2C%20du%20%5C%2C%20dv)
Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.
![\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%20u%20%3D%20r%5Ccos%28%5Ctheta%29%20%5C%5C%20v%20%3D%20r%5Csin%28%5Ctheta%29%20%5C%5C%20u%5E2%2Bv%5E2%20%3D%20r%5E2%5C%5C%20du%5C%2Cdv%20%3D%20r%5C%2Cdr%5C%2Cd%5Ctheta%5Cend%7Bcases%7D)
Then
![\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20768%20%5Cint_%7B-1%7D%5E1%20%5Cint_%7B-%5Csqrt%7B1-v%5E2%7D%7D%5E%7B%5Csqrt%7B1-v%5E2%7D%7D%20u%5E2%5C%2Cdu%5C%2Cdv%20%3D%20768%20%5Cint_0%5E%7B2%5Cpi%7D%20%5Cint_0%5E1%20%28r%5Ccos%28%5Ctheta%29%29%5E2%20r%5C%2Cdr%5C%2Cd%5Ctheta%20%5C%5C%5C%5C%20~~~~~~~~~~~~%20%3D%20768%20%5Cleft%28%5Cint_0%5E%7B2%5Cpi%7D%20%5Ccos%5E2%28%5Ctheta%29%5C%2Cd%5Ctheta%5Cright%29%20%5Cleft%28%5Cint_0%5E1%20r%5E3%5C%2Cdr%5Cright%29%20%3D%20%5Cboxed%7B192%5Cpi%7D)
Helen should of put 0 first, because if you're gonna divide, you need to put the product first, then one of the factoes.
So instead it should be 0÷(-7)=0
Answer:0.625 or 5/8
Step-by-step explanation:
Answer:
Sorry is there any more information i will then change answer
Step-by-step explanation: