In the coordinate plane, quadrilateral ABCD has vertices A(-2, 3), B(4,5), C(10, -1), and D(8, -9). Let E, F, G, and H
Aliun [14]
Answer:
G(3, 1), H(2, 3)
Step-by-step explanation:
When D is the midpoint of EG, it means ...
D = (E + G)/2
or
G = 2D -E = 2(3,4) -(3,7) = (2·3-3, 2·4-7) = (3, 1)
Likewise, H is ...
H = 2D -F = 2(3,4) -(4,5) = (2·3-4, 2·4-5) = (2, 3)
not my answer but i hope it helps you.
Estimate 95 to 90 and 8 to 10
now multiply 90 and 10 = 900
The answer is : multiply 5
(a)
The inverse is when you swap the variables and solve for y.
g(t) = 2t - 1 (Note: g(t) represents y)
rewrite as: y = 2t - 1
swap the variables: t = 2y - 1
solve for y: t + 1 = 2y

= y
Answer for (a):
=
(b)
Same steps as part (a) above:
h(t) = 4t + 3
rewrite as: y = 4t + 3
swap the variables: t = 4y + 3
solve for y:
Answer for (b):
= 
(c)

replace all t's in the

equation with

=

=

=

=
Answer for (c):
= 
(d)
h(g(t)) = h(2t - 1) = 4(2t - 1) + 3 = 8t - 4 + 3 = 8t - 1
Answer for (d): h(g(t)) = 8t - 1
(e)
h(g(t)) = 8t - 1
y = 8 t - 1
t = 8y - 1
t + 1 = 8y

= y
Answer for (e): inverse of h(g(t)) =
Question is: how many 84s will fit in 5376? Let's think about some easy multiples:
84 * 100 = 8400, so it's too big
84 * 10 = 840, so it might work
84 | 5376 | 10
-840
84 | 4536 | 10
-840
84 | 3696 | 10
-840
84 | 2856 | 10
-840
84 | 2016 | 10
-840
84 | 1176 | 10
-840
84 | 336
We can't fit any more 840 in 336, so we check how many 84s are in 336 and what's the remainder:
84 | 336 | 4
- 336
So there's no remainder. Now we add all the partial quotients to get the final result:
10 + 10 + 10 + 10 + 10 + 10 + 4 = <u>64
</u>It's correct, I checked it with calculator. I just hope you'll be able to read something from that, it's quite difficult to do partial dividing with no pencil and paper :)