Answer:
For this answer, I will label the points. Starting at the top left, then top right, then bottom left and bottom right let the points be A, B, C, D.
The new coordinates will be
A(-4,10)
B(4,10)
C(-4,4)
D(4,4)
Step-by-step explanation:
The question is asking for a dilation which is a transformation that makes an image proportionately smaller or larger by a scale factor. The scale factor is how much smaller or larger the shape will be, if the scale factor is between 0 and 1 then it will shrink, if it is greater than one then the image will stretch (be larger). In this case, the scale factor is 2, therefore the image will stretch. Since the center of dilation is the origin, to find the new coordinates simply multiply each x and y value by the scale factor. So A's original coordinates (-2,5) become (-4,10) and so forth. Therefore the equation for this dilation is (x, y) → (2x,2y).
<h3>Sample space = {a,b,c,d,e,f}</h3><h3>Event space = {a,c}</h3>
We simply list all of the letters mentioned as they are the possible outcomes. We can only pick one item from the sample space. The event space is the set of outcomes where we want to happen (picking either an 'a' or 'c').
Answer:
Q=(0.4.-1.7)
R=(2.4. 9.3)
S=(-10.6. 7.3)
Step-by-step explanation:
Just did it and got it wrong and these are the correct answers
Answer: sin u = -5/13 and cos v = -15/17
Step-by-step explanation:
The nice thing about trig, a little information goes a long way. That’s because there is a lot of geometry and structure in the subject. If I have sin u = opp/hyp, then I know opp is the opposite side from u, and the hypotenuse is hyp, and the adjacent side must fit the Pythagorean equation opp^2 + adj^2 = hyp^2.
So for u: (-5)^2 + adj^2 = 13^2, so with what you gave us (Quad 3),
==> adj of u = -12 therefore cos u = -12/13
Same argument for v: adj = -15,
opp^2 + (-15)^2 = 17^2 ==> opp = -8 therefore sin v = -8/17
The cosine rule for cos (u + v) = (cos u)(cos v) - (sin u)(sin v) and now we substitute: cos (u + v) = (-12/13)(-15/17) - (-5/13)(-8/17)
I am too lazy to do the remaining arithmetic, but I think we have created a way to approach all of the similar problems.
Answer:think of many situations you can come up with or problems you can solve which will end up with the same answer if u use different methods
Step-by-step explanation:
