Answers:
Upper Limit = 2.55 cc
Lower Limit = 2.25 cc
cc stands for "cubic centimeters".
==========================================
Explanation:
The tolerance tells us how much we can go over or under our target. It's our margin of error. We add and subtract the tolerance (0.15) from the target we're trying to aim for (2.4)
So the lowest dosage allowed is 2.4-0.15 = 2.25 cc
The highest dosage allowed is 2.4+0.15 = 2.55 cc
The upper limit is 2.55 cc while the lower limit is 2.25 cc
On a number line, we would find that 2.4 is at the midpoint of 2.25 and 2.55; so its at the center of those lower and upper bounds. The distance from the center to each endpoint is the same. That distance would be the tolerance 0.15
Answer:
=ba+2a^2-a
Step-by-step explanation:
The formula is
A=p (1+r/k)^kt
A amount of the withdrawal?
P present value 4000
R interest rate 0.06
K compounded monthly 12
T time 2 years
A=4,000×(1+0.06÷12)^(12×2)
A=4,508.64
So it's c
<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?
Use Hooke's law... (just kidding)
Break down each force vector into horizontal and vertical components.
The resultant force is the sum of these vectors,
and has magnitude
The closest answer is D.
