Answer:
D. the incenter
Step-by-step explanation:
Hope this helps
Remember
(x^m)/(x^n)=x^(m-n)
7^16/6^12=7^(16-12)=6=7^4
so
7^4=7^-18/?
?=7^?
7^4=7^-18/7^?
7^-18/7^?=7^(-18-?)=7^4
-18-?=4
add 18
-?=22
tmes -1
?=-22
7^-22
answer is C
<h3>
Answer: The same or equal</h3>
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Explanation:
Consider an example like 3, 6, 12, 24, 48, ...
The ratio between consecutive terms is
- 6/3 = 2
- 12/6 = 2
- 24/12 = 2
- 48/2 = 2
Each time we divide any given term over its previous one, we get the same ratio 2. We call this the common ratio. In terms of notation, the variable r is used for the common ratio, so r = 2 in this case.
As another example, the geometric sequence 5, 50, 500, 5000, ... has r = 10 as the common ratio because we multiply each term by 10 to get the next one. Moving forward has us multiply by r, moving backward and we divide by r. The value of r cannot be zero, but it can be negative.
An example with r being negative would be something like
1, -1, 1, -1, 1, -1, ....
we just bounce back and forth between those two values. In this case, r = -1.
<span>Vector Equation
(Line)</span>(x,y) = (x,y) + t(a,b);tERParametric Formx = x + t(a), y = y + t(b); tERr = (-4,-2) + t((-3,5);tERFind the vector equation of the line passing through A(-4,-2) & parallel to m = (-3,5)<span>Point: (2,5)
Create a direction vector: AB = (-1 - 2, 4 - 5)
= (-3,-1) or (3,1)when -1 (or any scalar multiple) is divided out.
r = (2,5) + t(-3,-1);tER</span>Find the vector equation of the line passing through A(2,5) & B(-1,4)<span>x = 4 - 3t
y = -2 + 5t
;tER</span>Write the parametric equations of the line passing through the line passing through the point A(4,-2) & with a direction vector of m =(-3,5)<span>Create Vector Equation first:
AB = (2,8)
Point: (4,-3)
r = (4,-3) + (2,8); tER
x = 4 + 2t
y = -3 + 8t
;tER</span>Write the parametric equations of the line through A(4,-3) & B(6,5)<span>Make parametric equations:
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in -3
-3 = 5 + 4t
(-8 - 5)/4 = t
-2 = t
For y sub in -8
-8 = -2 + 3t
(-8 + 2)/3 = t
-2 = t
Parameter 't' is consistent so pt(-3,-8) is on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (-3,-8) on the line?<span>Make parametric equations:
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in 1
-1 = 5 + 4t
(-1 - 5)/4 = t
-1 = t
For y sub in -7
-7 = -2 + 3t
(-7 + 2)/3 = t
-5/3 = t
Parameter 't' is inconsistent so pt(1,-7) is not on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (1,-7) on the line?<span>Use parametric equations when generating points:
x = 5 + 4t
y = -2 + 3t ;tER
X-int:
sub in y = 0
0 = -2 + 3t
solve for t
2/3 = t (this is the parameter that will generate the x-int)
Sub t = 2/3 into x = 5 + 4t
x = 5 + 4(2/3)
x = 5 + (8/3)
x = 15/3 + (8/3)
x = 23/3
The x-int is (23/3, 0)</span>What is the x-int of the line r = (5,-2) + t(4,3); tER?Note: if they define the same line: 1) Are their direction vectors scalar multiples? 2) Check the point of one equation in the other equation (LS = RS if point is subbed in)What are the two requirements for 2 lines to define the same line?
