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\(x^2-ax+2016=0\)

fixed your LaTeX

(it was kind of weird to read)

a) Let $x be the price of a ticket.

Price increase = $(x - 20).

For $1 increase in price, 15 fewer tickets are sold.

For $1(x - 20) increase in price, 15(x - 20) fewer tickets are sold.

Therefore if the price is $x, then the number of tickets sold would be 750 - 15(x - 20), which is 1050 - 15x upon simplification.

The revenue is equal to (number of tickets sold) * (price of each ticket), which is $x(1050 - 15x) = $15x(70 - x).

You can find the maximum revenue by completing the square.

b) is just similar.

i didn't figure out the question but maybe try finding patterns

hope that helps 

maybe this will help but as u can see in the url the numbers are different 

https://www.quora.com/The-sum-of-two-numbers-is-8-and-their-product-is-15-What-is-the-sum-of-their-reciprocals

Thanks for clarifying, loki.

That is indeed what I meant.

(oops next time I'll do parentheses!)

Because AB is a diameter, by Thale's theorem,

\(\angle ACB = \angle ADB = 90^\circ\)

That means \(\angle CAB = \arccos\left(\dfrac24\right) = 60^\circ\) by simple trigonometry.

Also, because \(\angle ADB = 90^\circ\), considering the interior angle sum of \(\triangle ADB\),

\(\angle DBA = 45^\circ\). This means \(\triangle ADB\) is a right isosceles triangle, which means AD = DB.

By Pythagorean theorem, we get \(AD = DB = 2\sqrt 2\).

Now consider \(\triangle ADC\) and use Law of Cosine. I believe you can finish it from here on your own.

maybe this will help but it only involves when no 2 adjacent houses can be the same color:

https://ncona.com/2017/02/house-painting-problem/

The key to this problem is realizing that factorials past 5! have a units digit of 0. (Make sure you see why)

Therefore, we only have to look at 2! + 3! + 4! = 2 + 6 + 24 = 32

So the units digit is 2.

just multiply each case and divide by the repeating numbers

does that help?

it might not help if you don't know a lot about counting and probability.

What is the definition of P? Is it the center of the circumscribing circle?

That means \(\begin{cases}x - y = \pm4 \\y - z = \pm5 \\ z - w = \pm7\end{cases}\)

Let \(a = x-y, b = y- z, c = z- w\)

\(x - w = a + b + c\)

We can interpret the question as finding the minimum sum over a, b and c.

By trying all 8 possible combinations, we get the smallest possible value to be |4 + 5 + (-7)| = 2

Thanks a lot!

To get the largest amount, we want to multiply two large amounts. 

We only have one times sign. 

To maximize a factor, we put parenthesis here: 3*(2 + 8/2 + 2)

This equals 3*8=24.

So the largest value one can obtain is 24.

Treat BC as the base, and the height would be the perpendicular distance from A to BC.

Distance from A to BC = 8 - 2 = 6.

BC = 7 - 4 = 3

Then, we use the formula \(A = \dfrac{bh}{2}\) to determine the area. 

You can try to finish it from here.

1883 / 3 = ~ 628. Take its square root and that would be the middle number of the three.

Sqrt(628) =about 25 - this would be the middle number. Since they are 3 consecutive ODD numbers, then:

23^2 + 25^2 + 27^2 =1883

We first find the radius,

Let r be the radius.

\(\dfrac{40^\circ}{360^\circ} \cdot \pi r^2 = 4\pi\\ r = 6\)

Then we can use the formula \(P = \dfrac{\theta}{360^\circ} \cdot \left(2\pi r\right) + 2r\) to calculate the perimeter of the sector.

I believe you can finish it on your own.

In order for a number to be divisible by 6, it needs to be divisible by both 3 and 2. 

The divisiblity rule for 2 is if the end digit is an even number.

The divisiblity rule for 3 is if all the digits added up together can be divided by 3.

Even numbers are easy to spot, so let's focus on divisibility by 3.

5+2+2+8+X = 3n 

17+X=3n

Now, we can use trial and error.

Case 1: n=6

17+X=18

X=1, which is not an even number, so discard this solution.

Case 1: n=7

17+X=21

X=4, which is an even number, so this is a valid solution.

At this point, we don't have to test for n=8 because the question states that there is only one digit for X.

So the answer is X=4.

"deleted"

Since are looking for a cubic polynomial, let it be ax^3 + bx^2 + cx + d = 0.  So

\(a(\sqrt[3]{2} + \sqrt[3]{4})^3 + b(\sqrt[3]{2} + \sqrt[3]{4})^2 + c(\sqrt[3]{2} + \sqrt[3]{4}) + d = 0\)

This expands to

\(a (2 + 3 \sqrt[3]{2^2} \sqrt[3]{4} + 3 \sqrt[3]{2} \sqrt[3]{4^2} + 4) + b(\sqrt[3]{2^2} + 2 \sqrt[3]{2} \sqrt[3]{4} + \sqrt[3]{4^2}) + c(\sqrt[3]{2} + \sqrt[3]{4}) + d = 0\)

Expanding everything out, and comparing the coefficients, we get

a + 3b + 3c + d = 16,
-6b + 3c + 3d = -24,
c - 3d = 22,
d = -6.

The solution to this system is a = 1, b = 3, c = 4, d = -6, so the cubic is x^3 + 3x^2 + 4x - 6.

x - y = 8*sqrt(7)

I have edited it

Draw square(ABCD).

On the diagonal AC, find point E such that AE = 1.

Find point X on BC such that EX is perpendicular to CB.

Since AC is a diagonal of a square whose sides are each 1, the length of AC = sqrt(2).

Since AE = 1, EC = sqrt(2) - 1..

Triangle(ACB) is similar to triangle(ECX)      [by AA].

This makes                       CE/CA = EX/AB

--->              (sqrt(2) - 1) / sqrt(2)  =  EX / 1

--->              (sqrt(2) - 1) / sqrt(2)  =  EX                  

--->                                         EX  =  0.293  (approximately)

[It would be the same istance to side CD.]  

Thank you !

if \(\overline{TR} = \overline{TI}\) then \(\angle R = \angle I\) all angles of a triangle add up to 180 degrees so 180 - 52 - 52 = 76 so \(\angle T\) = 76 degrees and that is all I can do.