a) Show that the locus of points specified by 𝑥 = 3 cos 𝜃 + 4 and 𝑦 = 3 sin 𝜃 − 2 is a circle.
cos θ = (x - 4) /3 sin θ = (y + 2) / 3
And
cos^θ + sin^2θ = 1 ....so.....
(x - 4)^2 / 3^2 + (y + 2) /3^2 = 1
(x - 4)^2 + (y + 2)^2 = 3^2
(x -4)^2 + (y + 2)^2 = 9
This is a circle cenrered at ( 4, -2) with a radius of 3
y = -2x + 8 would work
In fact.....any equation in the form y = -2x + b would work
xy = 56 ⇒ y = 56/x
7 (1/x) + 14(1/y) = 4
7/x + 14/y = 4
7/x + 14 / (56/x) = 4
7/x + (14/56)x = 4
7/x + (1/4) x = 4 multiply through by 4x
28 + x^2 = 16x rearrange as
x^2 - 16x + 28 = 0 factor as
(x - 14) ( x - 2) = 0
Set each factor to 0 and solve for x and we have that
x = 14 or x = 2
And y =
56/14 = 4 or 56/2 = 28
So (x,y) = (2, 28)
Just calculate 38 x 8
I would go in this order : Algebra, Pre-Cal,Trig
Can't tell you how long it would take......but maybe a year
2^2 ( 1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12) = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44,48
3^2(1,2,3, 5) = 9, 18, 27, 45
5^2(1, 2) = 25, 50
7^2 = 49
19 numbers
Given is the function f(x) = -x2
a. The graph of f shifted 3 units down. The graph that comes into existence, belongs to the function g. Calculate the points of intersection of the graph of g with the line y = -28
The new function is
g(x) = -x^2 - 3 set this = to 28
-x^2 - 3 = -28 add 3 to both sides
-x^2 = - 25 multiply through by - 1
x^2 = 25 take both roots
x = ±√25
x = ±5
So....the points of intersection are (5, -28) and (-5, 28)
b. The graph of h starts from the graph of f and has been shifted up a few unit. Then the graph is multiplied with the factor 2. The function notation for the new graph is h(x) = -2x2 + 6. Calculate how many units graph of f was shifted up to form the new graph.
Let the sfift of f be c units
So we have that f(x) - -x^2 + c
2f(x) = 2 [-x^2 + c ] = -2x^2 + 6
-2x^2 + 2c = - 2x^2 + 6
So
2c = 6
c = 3 units
Also could you give me like which to start with first?
\(\frac{269278645298734569827634598264956384775628456298743}{9865298746528945692873659827465245}\) = 27295538859730818.3446586092722013
Nunya
x^(1/3)
Put parentheses around the 1/3 so it knows it is all in the exponent
1, 4, 9, 12, 16, 18, 20, 24, 27, 28, 32, 36, 40, 44, 48=15 have perfect squares as factors.
P.S. 1, 4, 9 have the themselves as "pefect squares as factors".
Solve for t: (t + 1) (3 t - 3) - (t + 2) (3 t - 5) = 2
Expand out terms of the left hand side: 7 - t = 2
Subtract 7 from both sides: -t = -5
Multiply both sides by -1: t = 5
deleted
The volume of a cube with side S is S3
All three sides are the same so this could be written
as Length times Width times Height VC = L*W*H = 5*5*5
The volume of a pyramid is the area of the base
times one-third of the height so this could be written
as Length times Width times Height divided by 3 VP = L*W*H / 3 = 5*5*H/3
So, simply by observation, for the volumes to be equal,
the Height of the pyramid has to be 3 times the height
of the cube HP = 3 * 5 inches = 15 inches
.
Thanks!
There are two positive integers c for which the equation \(5x^2 + 11x+c=0\) has rational solutions.
What is the product of those two values of \(c\)?
\(\begin{array}{|rcll|} \hline \mathbf{ 5x^2 + 11x+c} &=& \mathbf{0} \\\\ x &=& \dfrac{-11\pm \sqrt{121-4\cdot 5c} }{2\cdot 5} \\ \hline 121-4\cdot 5c &>& 0 \\ 121-20c&>& 0 \\ 121&>&20c \\ \mathbf{\dfrac{121}{20}} &>& \mathbf{c} \\ \hline \end{array}\)
The possible values of positive integers \(c\) are \(\{1,2,3,4,5,6\}\)
\(\begin{array}{|c|l|c|} \hline c & \sqrt{121-20c} & \text{rational solutions} \\ \hline \mathbf{6} & \sqrt{121-20\cdot 6}=\sqrt{1}= 1 & \checkmark \\ \hline 5 & \sqrt{121-20\cdot 5}=\sqrt{21} & \\ \hline 4 & \sqrt{121-20\cdot 4}=\sqrt{41} & \\ \hline 3 & \sqrt{121-20\cdot 3}=\sqrt{61} & \\ \hline \mathbf{2} & \sqrt{121-20\cdot 2} =\sqrt{81} = 9 & \checkmark\\ \hline 1 & \sqrt{121-20\cdot 1}=\sqrt{101} & \\ \hline \end{array}\)
So \(c=6\) and \(c=2\) and \(2\cdot 6 = 12\)
\(\frac{\left( x + \dfrac{1}{x} \right)^6 - \left( x^6 + \dfrac{1}{x^6} \right) - 2}{\left( x + \dfrac{1}{x} \right)^3 + \left( x^3 + \dfrac{1}{x^3} \right)} \)
Just by inspection I can see that as x moves away from 1 in either direction the function value will increase.
So the minimum will occur when x=1
that minimum will be
\(\frac{2^6-2-2}{2^3+2}=\frac{60}{10}=6\)
Just as our guest said,
Suppose that \(2x^2 - 5x + k = 0\) is a quadratic equation with one solution for \(x\).
Express \(k\) as a common fraction.
\(\begin{array}{|rcll|} \hline \mathbf{2x^2 - 5x + k} &=& \mathbf{0} \\\\ x &=& \dfrac{5\pm \sqrt{25-4\cdot 2k} }{2\cdot 2} \\ \hline 25-4\cdot 2k &=& 0 \\ 25-8k &=& 0 \\ 8k &=& 25 \\ \mathbf{k} &=& \mathbf{\dfrac{25}{8}} \\ \hline \end{array}\)
If \(-5≤a≤-1\) and \(1≤b≤3\), what is the least possible value of \(\left(\dfrac{1}{a}+\dfrac{1}{b} \right) \left(\dfrac{1}{b}-\dfrac{1}{a} \right)\)?
1.)
\(\left(\dfrac{1}{a}+\dfrac{1}{b} \right) \left(\dfrac{1}{b}-\dfrac{1}{a} \right) = \dfrac{1}{b^2}-\dfrac{1}{a^2}\)
2.)
\(\begin{array}{lrcll} &\mathbf{ -5}\le &\mathbf{a}& \le \mathbf{-1} \quad | \quad square \\ & 25\le &a^2& \le 1 \\ (1) & \mathbf{\dfrac{1}{25}} \ge & \mathbf{\dfrac{1}{a^2}} & \ge \mathbf{1} \\ \hline & \mathbf{1} \le &\mathbf{b}& \le \mathbf{3} \quad | \quad square \\ & 1 \le &b^2& \le 9 \\ (2)& \mathbf{1} \ge & \mathbf{\dfrac{1}{b^2}} & \ge \mathbf{\dfrac{1}{9}} \\ \hline (2)-(1): & \mathbf{1} -\mathbf{\dfrac{1}{25}} \ge & \mathbf{\dfrac{1}{b^2}} -\mathbf{\dfrac{1}{a^2}} & \ge \mathbf{\dfrac{1}{9}} -\mathbf{1} \\ & \dfrac{24}{25} \ge & \dfrac{1}{b^2} - \dfrac{1}{a^2} & \ge -\dfrac{8}{9} \\ \end{array}\)
The least possible value is \(\mathbf{-\dfrac{8}{9}}\)
if x + y + z = 1 and x^2 + y^2 + z^2 = 3, then find the value of xy + xz + yz.
Expand (x + y + z)^2
\(\begin{array}{rcll} (x + y + z)^2 &=& 1 \\ x^2+y^2+z^2+2(xy+yz+xz) &=& 1 \quad | \quad x^2 + y^2 + z^2 = 3 \\ 3+2(xy+yz+xz) &=& 1 \\ 2(xy+yz+xz) &=& -2 \\ \mathbf{xy+yz+xz} &=& \mathbf{-1} \\ \end{array}\)
x * y = 56..............................(1)
7*1/x + 14*1/y = 4...................(2)
Can you solve the 2 simultaneous equations? Try it.
Positive integers x and y have a product of 56 and x
