To make writing easier, replace √5
by x in the left side of the original expression. After replacing, you will get this polynomial (3+x)*3*(5+3x))^3.
(3 + x)*3*(5 + 3x))^3 = 81x^4 + 648x^3 + 1890x^2 + 2400x + 1125. (*)
Now, to get first integer number of the right side, I consider the terms of this polynomial (*) with even degrees
even(x) = 81x^4 + 1890x^2 + 1125
and substitute there x2 = 5.
I get then "second integer number in the answer" = 648*5 + 2400 = which is easy to compute = 5640. Therefore, the final ANSWER is 12600 + 5640* √5 (**)
Finally, to check the answer, I calculated (**) and compared with the left side of the original expression. I got the same number, which confirms correctness of my solution.
Suppose the total number of 1 ft2 squares is A = LW
We don't want to count the 4 corner squares twice, so since the length is L, Then there are 2(L-2) non-corner squares along the 2 lengths and 2(W-2) non-corner squares along the 2 widths. So there are 2(L-2)+2(W-2) + 4 corner squares. That's 2L-4+2W-4+4 = 2L+2W-4 squares.
2/7 * LW = 2L * 2W - 4
2LW = 14L + 14W - 28
LW = 7L + 7W - 14
LW = 7L - 7W = -14
Complete the rectangle. We want to find what we must add to both sides so that LW-7-7W will factor in the form
Multiply that out: LW + qL + pW + pq
We see q=-7, p=-7, pq=49
Now we have to add 49 to both sides
LW - 7L - 7W+49=- 14+49
(L - 7) * (W - 7) = -14 + 49
(L - 7) * (W - 7) = 35
The factors of 35 are
35x1 = 7x5 Two possibilities if L > W
L-7=35, W-7=1 or L=42, W=8, LW = 336
L-7=7, W-7=5 or L=14, W=12, LW = 168
The smallest possible area is 168 ft2.
the coordinate of point x is -18 1/2
the coordinate of point is y is 25 1/3
-18 1/2.......................................................................................................25 1/3
the distance between X and Z is one-third of the distance between Z and
-18 1/2......................................................................................................25 1/3
the distance between X and Y
d = the distance between and Y divided by 4
d = (25 1/3 + 18 1/2)/4
d = 263/24
3d = (3 * 263)/24 = 263/8
the coordinate of the point Z is -18 1/2 + 263/24 = -181/24 = -7 13/24
25 1/3 - 263/8 = -181/24 = -713/24
answer: the coordinate of the point Z is -7 13/24
The normal has an equation
6y + 2x = -1 (given).
Its slope is -1/3.
It means that the tangent line to the parabola has the slope m= 3. From the given equation of the parabola, the slope of the tangent line at abscissa x is (1/2) * 3 * 2x = 3x (after taking the derivative).
In order for the slope of a tangent line 3x be m= 3, x should be equal to 1.
Thus we found out that x-coordinate of point A is x= 1.
Substitute x= 1 into equation of the normal line 6y + 2x = -1 . You will get 6y + 2*1 = -1, or 6y = -1 - 2 = -3, which gives y = -3/6 = -1/2.
Thus we learned that tangent point A on the parabola has coordinates A = (x,y) = (1, -1/2).
Now we know that the parabola passes throw the point (1, -1/2).
From equation of the parabola, it means that
-1/2 = (1/2) * (3 * 12 - k).
-1 = 3 - k,
k = 3 + 1 = 4.
k = 4.
Part ( i ) is completed in full.
Part ( ii ) is simple arithmetic.
Imagine that the cells of the 2x2 grid are denoted in this way
There are 2 basic cases: (a) A2 and B1 are of different colors (and different from A1 and B2),
(b) A2 and B1 are of the same colors (and different from A1 and B2)
Consider case (a) first.
(a) For A1, I can use any of 6 colors;
then for A2 I can use any of remaining 5 colors;
for B1 I can use any of remaining 4 colors;
for B2 I can use (and I must use) one of remaining 3 colors OR the same color as A1.
In all, I have 6*5*4*(3+1) = 6*5*4*4 = 480 different colorings in case (a).
Consider case (b) next.
(b) For A1, I can use any of 6 colors;
then for A2 I can use any of remaining 5 colors;
for B1 I use the same color as A2, so I have no choice;
for B2 I can use (and I must use) one of remaining 4 colors OR the same color as A1.
In all, I have 6*5*1*(4+1) = 6*5*1*5 = 150 different colorings in case (b).
Cases (a) and (b) are DISJOINT, so the ANSWER to the problems' question is this sum 480 + 150 = 630.
If in a right angled triangle ABC, the altitude h = BD from the right angle B to the hypotenuse AC divides the hypotenuse by segments of the length x and y, then
h2 = x*y. (1)
It is the general formula and the general statement/fact from Euclidean geometry. In our case h = BD = 3, x = AD = 9, and the problem asks about y = DC. From formula (1), y = h2/x = 32/9 = 9/9 = 1.
DC = 1.
The point (x,y) is in quadrant III, because x and y are both negative.
Because the point is 6 units from the x-axis, y is -6. So the point is (x,-6).
The distance from (x,-6) to (5,7) is 15. The difference in the y coordinates is 13, the distance is 15; find the difference in the x coordinates.
132 + b2 = 152
169 + b2 = 225
b2 = 56
b = √56
The point (x,y) is √56 units left of x=5, so the point is (5-sqrt(56)),-6)
The point is √n units from the origin:
(5 - √56)2 + 62 = (√n)2 = n
25 + 56 - 10√56 + 36 = n
n = 117 - 10√56 = 42.167 to 3 decimal places.
ANSWER: (approximately) 42.167