Solve 180=10x+10 to find your answer (:
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:
Angle 1 is congruent to angles 3, 5, and/or 7
Angle 2 is congruent to angles 4, 6, and/or 8
Angle 5 is congruent to angles 7, 3 and/or 1
Angle 6 is congruent to angles 8, 4, and/or 2
Any of these answers could work for the blanks.
Angles 1 and 3, 2 and 4, 5 and 7, and angles 6 and 8 are congruent because they are vertical angles. They have the same vertex. Not all of these are congruent to each other if this doesn’t make sense. It’s only 1 is congruent to 3, 2 congruent to 4, etc.
Then you have your corresponding angles. These are ones like angles 2 and 6, then 1 and 5. You can also have 8 and 4, or 7 and 3 as corresponding angles
Transversal angles are different. This would be like angles 3 and 4, or 1 and 2. They are not always congruent. The only time they will be congruent is if they are both 90°. Transversal angles are essentially supplementary angles on the transversal line (the line that intersects through the set of parallel lines)
