Answer seven and eight please
1 answer:
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Answer: Refer to the diagram below
- P ' is at (-7, 9)
- Q ' is at (1, 9)
- R ' is at (-6, 4)
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Explanation:
The diagram shows how point R moves to R'. We move up 2 units going from R(-6,0) to (-6,2). This lands us on the line of reflection. Then we move another 2 units up to land on (-6,4) which is the location of point R'.
The other points P and Q follow the same idea. Though the distances will be different from R. For P and Q, we'll move 7 units up to arrive at the line of reflection, then another 7 units to arrive at the proper locations of P' and Q', which are (-7,9) and (1,9) respectively.
Surface area of the square pyramid = 217.8 in²
Solution:
Side length of the square = 9 in
Area of the square = side × side
= 9 in × 9 in
= 81 in²
Area of the square = 81 in²
Perimeter of the base = 4 × 9 in = 36 in
Slant height of the triangle = 7.6 in
Surface area of the square pyramid
= 217.8 in²
Surface area of the square pyramid = 217.8 in²
Answer:
Step-by-step explanation:
The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.
Let
be the two paths.
Recall that if we parametrize a path C as with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is
Given any two points P, Q we can parametrize the line segment from P to Q as
r(t) = tQ + (1-t)P with 0≤ t≤ 1
The parametrization of the line segment from (1,1,1) to (2,2,2) is
r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)
r'(t) = (1,1,1)
and
The parametrization of the line segment from (2,2,2) to
(-9,6,5) is
r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)
r'(t) = (-11,4,3)
and
Hence