Answer:
Pool A.
Step-by-step explanation:
This can be solved through ratios. You would write it as 
So, 
for the first one, and 
for the second. You might want to use a calculator for that part, but you could also change the gallons to cups, which gives you a much better number of 10.515. So, 
, and 
So, pool A is at a faster rate. If you want it in gallons per minute, it's by 0.01 (with the 1 repeating) gallons per minute. By cups, it's 0.17525 cups per minute.
(The reason I did not do this by cross multiplying and dividing ratios is because it gives you the exact same answer if you are finding it per one, as I am here with per one minute. You can do it this way, it just wastes a bit of time.)
Answer:
Step-by-step explanation:
x^2 + 2x + 1 = 17
x^2 + 2x = 17 - 1
x^2 + 2x = 16
x^2 + 2x + 1 = 16 + 1
(x + 1)(x + 1) = 17
(x + 1)^2 = 17
x + 1 = sqrt 17
x = -1 (+ - ) sqrt 17
x = -1 + sqrt 17 or x = -1 - sqrt 17 <===
<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?
Answer:
x 2 + 5 x + 6
Step-by-step explanation:
y = m x + c
y- intercept
x = 0
y = (-2) (0) + 3
=3
(x =0, y = 3)
<h3>slope -2 </h3><h3>y = -1 x + 3</h3>
Answer:
0-5
Step-by-step explanation:
The Range Is 0-5 Atleast what i understand Gl Passing!
