The correct answer for the question shown above is the second option, the option B, which is: B. <span>W'(2, 10), X'(2, 2), Y'(10, 2)
The explanation is shown below: As you can see, the original triangle has the following coordinates </span><span> W(1, 5), X(1, 1), and Y(5, 1); the triangle must be dilated by a common scale factor, so if you analize the option B, you can notice that the triangle was dilated by a scale factor of 2.</span>
Answer:
x = 110
y = 40
Step-by-step explanation:
m∡5 = 30° because it forms a linear pair with ∡6, which is 150°
x + y = 150 (angles 4 and 6 are alternate-interior and are congruent)
angles 3 and 6 are same-side interior and are supplementary
therefore: x -2y = 180-150
if x + y = 150 we can say that x = 150 - y
we can plug '150-y' for 'x' in x - 2y = 30
150-y - 2y = 30
-3y = -120
y = 40
x = 110
Since density is the ratio of mass to (in this case) area, we can find the mass of the triangular region
by computing the double integral of the density function over
:

The boundary of
is determined by a set of lines in the
plane. One way to describe the region
is by the set of points,

So the mass is




Answer:
a)
Step-by-step explanation:
X'(-6;9)
Y'(9,3)
Z'(-8,-6)
Answer:
1. TRUE
2.FALSE
3.FALSE
4.FALSE
5.TRUE
Step-by-step explanation: