The blue dot is equal to -5
To find the x-intercept, substitute in 0 for y and solve for x. To find the y-intercept, substitute in 0 for x and solve for y.
x-intercept: (9/4,0)
y-intercept: (0,−9)
I think the answer is a please tell me if it’s right
Answer:450
Step-by-step explanation:
first subctract then add
Missing information:
How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

Answer:

Step-by-step explanation:
Given




Express the given point P as a unit tangent vector:

Next, find the gradient of P and T using: 
Where

So: the gradient becomes:

![\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]](https://tex.z-dn.net/?f=%5Ctriangle%20T%20%3D%20%5B%28sin%20%5Csqrt%203%29i%20%2B%20%28cos%20%5Csqrt%203%29j%5D%20%2A%20%20%5B%5Cfrac%7B%5Csqrt%203%7D%7B2%7Di%20-%20%5Cfrac%7B1%7D%7B2%7Dj%5D)
By vector multiplication, we have:




Hence, the rate is: