The largest possible value of y is 1 + sqrt(3).
Thank you so much!
The ellipse is (x - 5)^2/2^2 + (y + 1)^2/3^2 = 1.
f(g (x) ) = log [ (x)^2 / ( x - 1)^2 - 2x / (x - 1) ] =
x^2 - 2x ( x - 1)
log ____________ = "b"
(x - 1)^2
x > y > z.
Thanks, heureka......good to see you back !!!!!
Quad 3
Another complex number
Quad 2
sin B = cos C = 4/7
f(0) = a/2
Find the inverse
y = a / ( x + 2) get x by itself
(x + 2) y = a
x + 2 = a / y
x = a / y - 2 "swap" x and y
y = a/x - 2 = the inverse = f-1
f-1 (3a) = a / ( 3a) - 2 = (1/3) - 2 = -5/3
So
f (0) = f-1 (3a)
a/2 = -5/3
a = 2 (-5/3) = -10/3
Help please!
\(\text{Let $\vec{AC}=\vec{a}$} \\ \text{Let $\vec{CB}=\vec{b}$} \\ \text{Let $\vec{CD}=(1-\lambda)\vec{b}$} \\ \text{Let $\vec{DB}=\lambda\vec{b}$} \\ \begin{array}{|rcll|} \hline \text{In}~ \triangle \text{ABC}: \\ \hline \dfrac{1}{2}\vec{OB} &=& \vec{a} + \vec{b} \\ \mathbf{ \vec{OB} } &=& \mathbf{2(\vec{a} + \vec{b})} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{In}~ \triangle \text{OAE}: \\ \hline \vec{OE} &=& \dfrac{1}{2}\vec{OB} + \dfrac{1}{2}\vec{a} \quad | \quad \mathbf{ \vec{OB}=2(\vec{a} + \vec{b})} \\\\ \vec{OE} &=& \dfrac{1}{2}\times 2(\vec{a} + \vec{b}) + \dfrac{1}{2}\vec{a} \\\\ \vec{OE} &=& \vec{a} + \vec{b} + \dfrac{1}{2}\vec{a} \\\\ \mathbf{ \vec{OE} } &=& \mathbf{\dfrac{3}{2}\vec{a} + \vec{b}} \\ \hline \end{array} \)
\(\text{Let $\vec{OD}=\mu\vec{OE}$} \\ \text{Let $\vec{OD}=\vec{OB}-\vec{DB}$} \\ \begin{array}{|rcll|} \hline \text{In}~ \triangle \text{ODB}: \\ \hline \vec{OD} = \mu\vec{OE} &=& \vec{OB}-\vec{DB} \\ \mu\vec{OE} &=& \vec{OB}-\vec{DB} \quad | \quad \vec{OE}=\dfrac{3}{2}\vec{a} + \vec{b},~\vec{OB}=2(\vec{a} + \vec{b}),~ \vec{DB}=\lambda\vec{b} \\\\ \mu \left(\dfrac{3}{2}\vec{a} + \vec{b}\right) &=& 2(\vec{a} + \vec{b})-\lambda\vec{b} \\\\ \dfrac{3}{2}\vec{a}\mu + \mu \vec{b} &=& 2\vec{a} + 2\vec{b}-\lambda\vec{b} \\\\ \vec{a}\underbrace{ \left(\dfrac{3}{2}\mu-2\right)}_{=0} &=& \vec{b}\underbrace{\left( 2-\mu-\lambda\right)}_{=0} \\\\ \hline \dfrac{3}{2}\mu-2 &=& 0 \\ \dfrac{3}{2}\mu &=& 2 \\ \mathbf{\mu} &=& \mathbf{\dfrac{4}{3}} \\\\ \vec{OD} &=& \mu\vec{OE} \quad | \quad \vec{OE}=\dfrac{3}{2}\vec{a} + \vec{b}\\ \vec{OD} &=& \dfrac{4}{3} \left(\dfrac{3}{2}\vec{a} + \vec{b} \right)\\ \mathbf{\vec{OD}} &=& \mathbf{2\vec{a} + \dfrac{4}{3} \vec{b}} \\ \hline 2-\mu-\lambda &=& 0 \\ \lambda &=& 2-\mu \quad | \quad \mathbf{\mu=\dfrac{4}{3}} \\ \lambda &=& 2-\dfrac{4}{3} \\ \lambda &=& \dfrac{6-4}{3} \\ \mathbf{\lambda} &=& \mathbf{\dfrac{2}{3}} \\\\ \vec{CD} &=& (1-\lambda)\vec{b} \\ \vec{CD} &=& \left(1-\dfrac{2}{3}\right)\vec{b} \\ \mathbf{\vec{CD}} &=& \mathbf{\dfrac{1}{3}\vec{b}} \\\\ \vec{DB} &=& \lambda\vec{b} \\ \mathbf{\vec{DB}} &=& \mathbf{\dfrac{2}{3}\vec{b}} \\ \hline \dfrac{\vec{CD}}{\vec{DB}} &=& \dfrac{\dfrac{1}{3}\vec{b}}{\dfrac{2}{3}\vec{b}} \\\\ \dfrac{\vec{CD}}{\vec{DB}} &=& \dfrac{\dfrac{1}{3}}{\dfrac{2}{3}} \\\\ \dfrac{\vec{CD}}{\vec{DB}} &=& \dfrac{1}{3} \times \dfrac{3}{2} \\\\ \mathbf{\dfrac{\vec{CD}}{\vec{DB}}} &=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}\)
First Box: Quadrant 3.
Second Box: Because any ni * mi = a negative real number, the answer is another complex number.
Third Box: Quadrant 2.
Assuming that the semi-circle borders the rectangle's width......the total fencing required is
Three sides of the rectangle + perimeter of a semi-circle with a radius of 35/2 = 17.5 ft
( 35 + 2 * 25) + pi (17.5) =
85 + 17.5 pi ≈
140 ft
when memorization is less efficient
The ratio is 5/6.
What is the remainder when 2 + 4 + 6 + ... + 100 + 102 + 104 is divided by 7?
2 + 4 + 6 + ... + 100 + 102 + 104 = 2(1 + 2 + 3 + ... + 50 + 51 + 52)
1 + 2 + 3 + ... + 50 + 51 + 52 = (52)(53)/2
2((52)(53)/2) = (52)(53)
52 mod 7 = 3
53 mod 7 = 4
52*53 mod 7 = 12
12 mod 7 = 5
A cool way to do 22^2-17^2 = (22+17)(22-17) = 39 * 5 = 195
:))
=^._.^=
0 = (a-6a)/3a
-5a = 0
a = 0.
Modulus = sqrt ( (-2)^2 + 8^2) = sqrt (68) = sqrt ( 4 * 17) = 2sqrt (17)
5 ( -2 + 8i) = -10 + 40i
Modulus = sqrt ( (-10)^2 + (40)^2 ) = sqrt [ 1700) = sqrt (100 * 17) = 10sqrt (17)
a positive real number
22^2 pi - 17^2 pi = (484 - 289) * pi = 195 pi
Perimeter: 2 * (l+w)
2*(35+25)
2*60
120 feet
I'm really sorry, but the answer is 6.
x^3 + 5x^3 = 6x^3
I have no idea why I thought 1 + 5 = 8.
Thank you for helping me i got the quesiton correct
2, 4, 6, 1, 3, 5, 0, 2, 4, 6... (in mod 7)
Total of 52 numbers.
Each cycle of 7 numbers is 21, or 0.
50th, 51th, 52th number added is our answer. 2 + 4 + 6 = 12
no it is not correct
sqrt(6^2 + (-4)^2)
Take it from here. :))