i have some problems with these attached file , need help as soon as possible need to make sure i did them correctly
$\displaystyle \cos 2x = 0.5 $
$\displaystyle 2x = 60^{\circ}$
$\displaystyle 2x = 60^{\circ} \pm 360 k$
$\displaystyle x = 30^{\circ} \pm 180 k$
$\displaystyle \mbox{Or } x = 150^{\circ} \pm 180k
$
1. cos 2x = 0.5
2x = 60 degrees
2x = 60 degrees \pm 360 k
x = 30 degrees \pm 180 k
Or x = 150 degrees \pm 180k
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2. Look at the RHS, where you have $\displaystyle cos^2 \theta $
Remember that $\displaystyle cos^2 = 1 - sin^2 $
Therefore:
$\displaystyle 4 sin^2 \theta - 2 sin \theta = 4 - 4 sin^2 \theta - 1$
$\displaystyle 8 sin^2 \theta - 2 sin \theta - 3 = 0$
By solving for theta we find:
$\displaystyle sin \theta = 0.75 \ OR \ sin \theta = -0.5 $
3. $\displaystyle A = \pi R^2 $
But Perimeter $\displaystyle P = 2 \pi R $
$\displaystyle 20 = 2 \pi R $
$\displaystyle \frac{10}{\pi} = R $
Set R into A
$\displaystyle A = \pi (\frac{10}{\pi})^2 $
$\displaystyle A = \frac{100}{\pi} $
Now my area formula does not seem at all correct. See if you can figure out where i made a mistake.
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Now lets ASSUME we did get the right formula.
To maximise, we must find the derivative and equal it to 0
So dA/dr = 0
$\displaystyle 25 - (r - 5)^2 = 0 $
$\displaystyle 25 - r^2 + 10r - 25 = 0 $
$\displaystyle r^2 + 10r = 0 $
$\displaystyle r(r + 10) = 0 $
Therefore $\displaystyle r = 0 \ or \ r = -10 $
Set both 0 and -10 into the ORIGINAL equation
If r = 0
Area = 0
If r = -10
Area = -200
For the Architect Sum:
a) $\displaystyle 4x + 4y + x \pi $
b) $\displaystyle 4xy + \frac{1}{2} x^2 \pi $
c) $\displaystyle 100 = 4x + 4y + 2x \pi $
$\displaystyle y = 25 - (x + \frac{x \pi}{2}) $
Set y into the Area formula
Then we will eventually get: $\displaystyle A = 50x - 2x^2 + \frac{1}{2} x^2 \pi $
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Why cant i get these sums right??
Right lets assume the given values again
$\displaystyle \frac{dA}{dx} = 100 - 4x + x \pi = 0 $
$\displaystyle 100 = 4x - x \pi $
$\displaystyle 100 = x(4 - \pi) $
$\displaystyle x = \frac{100}{(4 - \pi)} $
Now just set $\displaystyle x $ into the ORIGINAL equation to find $\displaystyle y $
For the Graph sum:
a) To find where they intersect we must set y = y
$\displaystyle x^2 - 2x + 2 = -x + 4 $
$\displaystyle x^2 - x - 2 = 0 $
$\displaystyle x = 2 \ or \ x = -1 $
Set these values into an original equation.
Then the points are: y = -(2) + 4
And y = -(-1) + 4
So point P = (-1 ; 5)
Point Q = (2 ; 2)
b) For MPQ to be right-angled, the product of the two formulas must equal -1
I do not get the product = -1, so i am assuming that they are not right angled.
we have M(1 , 1), P(-1 , 5) and Q(2 , 2)
The slope of the line connecting P and Q is given by:
$\displaystyle m_{1} = \frac {2 - 5}{2 + 1} = \frac {-3}{3} = -1 $
The slope of the line connecting M and Q is given by:
$\displaystyle m_{2} = \frac {2 - 1}{2 - 1} = 1 $
since 1 is the negative reciprocal of -1, the line connecting P and Q is perpendicular to the line connecting M and Q. Thus this forms a right angle at Q. If we draw a line connecting M and P therefore, we form a right-triangle
Let's find the arc length first of all.
Recall:
$\displaystyle s = r \theta $
where s is the arc length and $\displaystyle \theta$ is the angle subtending the arc in radians
We are told that the perimeter is 20 cm this means that:
$\displaystyle P = r + r + s = 2r + r \theta = r(2 + \theta)$, where $\displaystyle P$ is the perimeter
$\displaystyle \Rightarrow r(2 + \theta) = 20 \Rightarrow 2 + \theta = \frac {20}{r} \Rightarrow \theta = \frac {20}{r} - 2$
Now, the area A of the sector will be given by:
$\displaystyle A = \frac {1}{2} r^2 \theta$
$\displaystyle = \frac {1}{2} r^2 ( \frac {20}{r} - 2) $
$\displaystyle = r^2 ( \frac {10}{r} - 1)$
$\displaystyle = 10r - r^2$
By completing the square, you will get the desired result
Umm, you forgot to find dA/dr janvdl, you did the manipulations on A, which is why you had weird values. (The negative valus should have hinted to you that you did something wrong)
Now lets ASSUME we did get the right formula.
To maximise, we must find the derivative and equal it to 0
So dA/dr = 0
$\displaystyle 25 - (r - 5)^2 = 0 $
$\displaystyle 25 - r^2 + 10r - 25 = 0 $
$\displaystyle r^2 + 10r = 0 $
$\displaystyle r(r + 10) = 0 $
Therefore $\displaystyle r = 0 \ or \ r = -10 $
Set both 0 and -10 into the ORIGINAL equation
If r = 0
Area = 0
If r = -10
Area = -200
We wish to maximize A, so we must find A' and set it equal to zero, a value for r that causes this to happen is the value we are looking for.
From above, $\displaystyle A = 25 - (r - 5)^2$
$\displaystyle \Rightarrow A = 10r - r^2 $
$\displaystyle \Rightarrow A' = 10 - 2r $
Set $\displaystyle A' = 0 $ we get:
$\displaystyle 10 - 2r = 0 $
$\displaystyle \Rightarrow r = 5 $
Note, the formula for a is a downward opening parabola, all we needed to do was find the value of r that gives the vertex. we didn't even need calculus here, we could have found this value by using:
$\displaystyle r = \frac {-b}{2a} $, where $\displaystyle a$ is the coefficient of $\displaystyle r^2$ and $\displaystyle b$ is the coefficient of $\displaystyle r$.
I assume you actually know how to draw the graphs, draw them both on the same pair of axis. see the graphs below.
a)
Note that we have the roots of $\displaystyle x^2 + e^x - 4 = 0$ where the graphs intersect (do you see why?). from the graphs we see that the intersections occur such that one is for a positive x-value and the other is for a negative x-value, thus we will have a positive and negative root.
b)
I can't really see the power of e very well, i think it is $\displaystyle x_{n}$ so let's run with that.
$\displaystyle x_{n+1} = - \sqrt {4 - e^{x_{n}}}$
$\displaystyle \Rightarrow x_{1} = - \sqrt {4 - e^{x_{0}}} = - \sqrt {4 - e^{-2}} = -1.96588$
$\displaystyle \Rightarrow x_{2} = - \sqrt {4 - e^{x_{1}}} = - \sqrt {4 - e^{-1.96588}} = -1.96468$
$\displaystyle \Rightarrow x_{3} = - \sqrt {4 - e^{x_{2}}} = - \sqrt {4 - e^{-1.96468}} = -1.96463$
so the approx for the negative root to three decimal places is -1.965
c)
the approximation will go towards the negative root, since within any reasonable range $\displaystyle \sqrt {4 - e^{x_{n}}$ will give a positive result. so with the minus on the outside, we will get a negative result
EDIT: So i believe i have taken care of all the problems janvdl didn't do or the ones he expressed doubt about, for the others, i think you're in good hands with him on the job, so i won't even check them
An angle in radians is defined to be
$\displaystyle \theta = \frac{s}{r}$
where s is the length of the arc the angle subtends in a cirlce of radius r.
For conversion purposes, you can use this formula to determine that there are $\displaystyle \pi$ radians in 180 degrees.
-Dan