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The area of the trapezoid ABCD is equal to 60. The length of one of of the bases is 7 units greater than the other base, and the height of trapezoid ABCD is equal to 5. What is the legth of the median of trapexoid ABCD ?

\text{For how many values of n with } 0\le n\le 100 \text{ is the graph of}  f(x) = \sin \left(x + n\right) \text{ identical to the graph of } g(x) = \cos x?

\(\text{For how many values of n with } 0\le n\le 100 \text{ is the graph of}\\ f(x) = \sin \left(x + n\right) \text{ identical to the graph of } g(x) = \cos x? \)

\( 0\le n\le 100 \\ 0\le n\le 31.83\pi \\ \)

\(sin0=cos(0+\pi/2)\quad \text{then it will work every 2pi after that}\\ n=\frac{\pi}{2},\;\;\frac{5\pi}{2},\;\;\frac{9\pi}{2},\;\;.....\frac{61\pi}{2}\\ 1,5,9,4c-3, 4*16-3\\ \text{so that appears to be 16 times.} \)

Here is the graph

As follows:

I think the picture is skewed a bit....I will assume the figures are rectangles

area of larger rectangle - area of smaller rectangle = shaded region

(3x+2)(5x)   -    (5x-2)(x) = shaded area

15x^2 + 10x   - 5x^2 + 2x

10x^2 + 12x   = shaded area

Alternatively, you could use Picks theorem for this particular figure.  See https://en.wikipedia.org/wiki/Pick%27s_theorem 

https://web2.0calc.com/questions/plz-help-asap_52#r1

Cut it into 3 triangles and a rectangle and add them together

triangle area = 1/2  b  h

   three triangles    1) 4 x 3      2)   4 x 3      3)  8 x 2

   rectangle     3 x 8

Is there any way to avoid sideways images?

Vertex -1,8    point   -3,0

Vertex form of parabola    y = a(x+1)^2 + 8    sub in point to calc 'a'

                                          0 = a(-3+1)^2 + 8      a = -2

y = -2 (x+1)^2 + 8

1.25 (x- 1/x)   = (x + 1/x)                                 in ratio 5:4     one side is 1.25 the other    1.25*4 = 5

x - 1/x          =   1.25x + 1.25/x

.25x = 2.25/x

.25x^2 = 2.25

.5x = 1.5

x = 3

Sorry,,, \(f(x) = \left\lceil\dfrac{1}{x+2}\right\rceil$ for $x > -2$, and $f(x) = \left\lfloor\dfrac{1}{x+2}\right\rfloor$ for $x < -2$. ($f(x)$ is not defined at $x = -2\)

I searched up, and this was actually posted before, and guest used the same answer both times. The other one has four thumbs down for guest which I do not clearly understand since guest must have spent a bit of effort into the answer. 

it's in radians

A Ford car left A to B and an Audi car left B to A at the same time. The speed of the Ford car to that of the Audi car was 4:3. The Ford decreased its speed by 25% and the Audi car increased its speed by 25% after they had passed each other. When the Ford car reached B, the Audi car was still 20 km away from A. Find the distance between A and B in km.

I have answered.

Here is a video on it

Let f(x)=12x+7 and g(x) = 5x+2 whenever x is a positive integer. Define h(x) to be the greatest common divisor of f(x) and g(x). What is the sum of all possible values of h(x)?

The gcd of f(x) and g(x) when x is a positive integer, is 1 for most values of x and 11 when x is 4, 15, 26, ...etc.

So the sum is 12.

Prove that there do not exist integers m and n such that

\(5m^2 − 6mn + 7n^2 = 2011\).

\(\begin{array}{|rclrcl|} \hline \mathbf{5m^2 - 6mn + 7n^2} &=& \mathbf{2011} \quad | \quad *5 \\\\ 25m^2 - 30mn + 35n^2 &=& 2011*5 \\ 25m^2 - 30mn + 9n^2+ 26n^2 &=& 2011*5 \\ (5m-3n)^2+ 26n^2 &=& 2011*5 \quad | \quad \mathbf{2011*5} \equiv \mathbf{6 \pmod{13}} \\ \hline \\ (5m-3n)^2+ 26n^2 &\equiv& 6 \pmod{13} \quad | \quad \mathbf{26n^2\pmod{13}} \equiv \mathbf{0} \\ (5m-3n)^2 +0 &\equiv& 6 \pmod{13} \\ (5m-3n)^2 &\equiv& 6 \pmod{13} \quad | \quad x= 5m-2n \\ \mathbf{x^2} &\equiv& \mathbf{6 \pmod{13}} \\ \hline \end{array}\)

The Quadratic Reciprocity Law with Legendre symbol:

Let \(\mathbf{p}\) be an odd prime. The integer \(\mathbf{a}\), prime to \(\mathbf{p}\), is said to
be a \(\mathbf{\text{quadratic residue}}\) or \(\mathbf{\text{nonresidue}}\) of \(\mathbf{p}\) according as the congruence

\(x^2 \equiv a\pmod{p}\)

is of is not solvable. The Legendre symbol \(\left(\dfrac{a}{p}\right)\) is defined to be \(+1\) or \(-1\)
according as \(\mathbf{a}\) is a quadratic residue or nonresidure of \(\mathbf{p}\).

\(\begin{array}{|rcll|} \hline \mathbf{x^2} &\equiv& \mathbf{6 \pmod{13}} \quad | \quad a = 6,\ p = 13 \\\\ \left(\dfrac{a}{p}\right) &=& \left(\dfrac{6}{13}\right) \\ \hline \\ \mathbf{\left(\dfrac{6}{13}\right)} &=& \left(\dfrac{2*3}{13}\right) \\\\ &=& \left(\dfrac{2}{13}\right)\left(\dfrac{3}{13}\right) \quad &| \quad \left(\dfrac{3}{13}\right)=\left(\dfrac{13}{3}\right) \\\\ &=& \left(\dfrac{2}{13}\right)\left(\dfrac{13}{3}\right) \\\\ &=& \left(\dfrac{2}{13}\right)\left(\dfrac{13\pmod{3}}{3}\right) \\\\ &=& \left(\dfrac{2}{13}\right)\left(\dfrac{1}{3}\right) \quad &| \quad \left(\dfrac{1}{3}\right) = 1 \\\\ &=& \left(\dfrac{2}{13}\right)*1 \\\\ &=& \left(\dfrac{2}{13}\right) \\\\ &=& \left(-1\right)^{ \frac{13^2-1}{8} } \\\\ &=& \left(-1\right)^{21} \\\\ &=& \mathbf{-1} \\ \hline \end{array}\)

\(\mathbf{\left(\dfrac{6}{13}\right)}=-1\), there do not exist integers m and n such that
\(5m^2 - 6mn + 7n^2 = 2011\).

You are missing a bracket.

I do not know what you are trying to ask. 

Does the 7 refer to the 7th root?

Thank you so much!~

Hello, Guest!

If radius   r = 4.25 then the diameter is 8.5?!?  It's not possible!!!! The diameter is over 12 units.

Angle (ACO) = arctan(3 / 12) = 14.03624347º

AD = [sqrt(3² + 12²)] / 2 = 6.184658438

AB = (AD) / cos(ACO) = 6.375

The radius of a circle is 6.375 units  

In the month of June, the temperature in Johannesburg, South Africa, varies over the day in a periodic way that can be modeled approximately by a trigonometric function.

The lowest temperature is usually around 3C, and the highest temperature is around 18C.

The temperature is typically halfway between the daily high and daily low at both 10 a.m., and 10 p.m.

and the highest temperatures are in the afternoon.

Find the formula of the trigonometric function that models the temperature T in Johannesburg t hours after midnight.

Define the function using radians.

\(\begin{array}{|rcll|} \hline \dfrac{18^\circ C + 3^\circ C}{2} &=& 10.5^\circ \\ \\ \dfrac{18^\circ C - 3^\circ C}{2} &=& 7.5^\circ \\ \\ \hline \mathbf{T} &=& \mathbf{10.5^\circ C+7.5^\circ C\cdot\sin\Bigg(\left(\frac{t+2}{12}-1\ \right)\cdot\pi \Bigg)} \\ \hline \end{array}\)

Image below with two squares, ABCD and BFGE, sharing a vertex.
Given that AE = 5, what is the length of DG?

My attempt:

\(\begin{array}{|rcll|} \hline DG^2 &=& 5^2+5^2 \\ DG^2 &=&2*5^2 \\ \mathbf{DG} &=& \mathbf{5\sqrt{2}} \\ \hline \end{array}\)