M = Tim's dollar amount
R = Rebecca's dollar amount
MR = Tim and Rebecca's dollars together
R = (M x 1/10) - 6
MR = M + ((M x 1/10) - 6)
MR = 70 + ((70 x 1/10) - 6)
MR = 70 + (7 - 6)
MR = 70 + 1
MR = 71
Answer:

Step-by-step explanation:
Starting from the top, the ant can only take four different directions, all of them going down, every direction has a probability of 1/4. For the second step, regardless of what direction the ant walked, it has 4 directions: going back (or up), to the sides (left or right) and down. If the probability of the first step is 1/4 for each direction and once the ant has moved one step, there are 4 directions with the same probability (1/4 again), the probability of taking a specific path is the multiplication of the probability of these two steps:

There are only 4 roads that can take the ant to the bottom in 2 steps, each road with a probability of 1/16, adding the probability of these 4 roads:

The probability of the ant ending up at the bottom is
or 0.25.
Answer:
Step-by-step explanation:
6k^2
------- = 2k; and:
3k
-15k
------- = -5
3k
-5 cannot be evenly divided by 3k, so there's a remainder instead.
6k^2 - 15k - 5
-------------------- = 2k -5 with remainder -5.
3k
21,600 v^3 been a while so not sure wether you need the sq symbol afterword let me know
Neither P, nor A are on the sketch
I guess P is the upper right corner of the rectangle
and A=(0,1)
P belongs to the line going through (1,0) and B(0,y)
<span>but we don't know the y-coordinate of B </span>
<span>the triangle is right and isosceles, so pythagoras a²+a²=2² ... 2a²=4 ... a²=2 ... a=sqrt2 </span>
now look at the right triangle BOA
<span>his hypotenuse is AB=sqrt2 and the <span>the kathete</span> OA is 1 </span>
so y²+1²=(sqrt2)² ... y²+1=2 ... y²=1.. y=1
so the coordinates of B are (0,1)
the line going through (1,0) and (0,1) is L(x)=-x+1
P belongs to this line, so the coordinates of P are P(x,-x+1) (0<x<1)
b) so if that's P, the height of the rectangle is -x+1 and the width=2x
<span>so its area A(x)=2x*(-x+1)= -2x²+2x
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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