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Ooh are you doing the intro to C&P course?

Lol I did it a while ago and remember being stuck on these card questions... I'll try to see if I can help...

Let's write an equation to solve this pronlem. 

First off, let's let x be the tens digit and the other be y. 

From the first of the problem, we have the equation

\(10x+y = 10y+x +72\)

From the second part of the problem, we have the equation

\(x=2y-1\)

Solving this system, we get that \(x=17,\:y=9\)

This equation cannot be satisfied. 

From the first equation, we have that \(y=x-8\)

There are only two possibilities. One is where x is 9 and y is 1, so we have 91. 

The other possibility is x is 8 and y is 0. 

Neither of these satisfy the conditions given.

Thanks! :)

40 * 6 * 75 * 12  =

40 * 6 *  [3 *25] * [ 3 *4 ]  =

40 * 6 * [ 3 * 3 ] * [ 25 * 4]  = 

6 * 40 *  9 *  100 =

54 * 4000  = 

54 * 4 * 1000  =

216 * 10^3

3 terminal 0's

Rewrite as rhe second and third equations

xz = x + z →  xz - x = z →  x(z -1) = z  →   x = z / (z -1)

yz = 2 (y+ z) → yz = 2y + 2z  →  yz - 2y = 2z →  y ( z -2)  = 2z →  y = (2z) / (z -2)

Rewrite the first equation. sub for x and y

[ z / ( z -1)]   [(2z) / (z -2)] =   z/ (z -1) + (2z) / ( z -2)

2z^2  =   z (z -2) + 2z (z -1)

2z^2 = 3z^2 - 4z

z^2 - 4z  = 0

z ( z - 4)  = 0

z = 0  (reject because it makes the second equation false )

z = 4

Rearrange as

6x^2 - 21x - 31 = 0

Using the Quad Formula, the largest x value is  [ 21 + sqrt [ 21^2 + 24 *31] ] / 12 ≈ 4.619    { as 3.141 found !!! }

a + 2b = -3  multiply through by -4 ⇒  -4a - 8b = 12   

Rearrange the second equation as  -4a + (2+m) b = k  

This system has infinite solutions when

2 + m = -8

m  = -10

And

k  =12

\(\text{By Vieta's formulae the sum of roots is -b/a, } \text{and the product is c/a}\)

if they are in an quadratic ax^2 + bx + c = 0

WE KNOW A = 1 :)

Therefore, the sum of the roots is -b and the product is c

The roots are 1 and -1, so the sum is 0 and the product is -1

Given that -b = 0 and c = -1, or b = 0 and c = 1, plug the values back into the equation x^2 + bx + c = 0 to get your final answer

Please like if this answer was correct. Thank you.

(x + 1) ( x -1)   =  x^2 + 0x - 1

\(\text{By graphing this on Desmos, you get -1.119 or 4.619, and we choose the bigger one, or 4.619} \)

The answer is \(0\)

I'm sorry, I used symbolab. 

We can write the equation in slope intercept form (y = mx +b) and convert it to Ax+By=C form.

m is the slope which is equal to (1-5)/(9-(-2)) = -4/11

y = -4x/11 + b

Substituting a point, we have 1 = -36/11 + b

b = 1 + 36/11 = 47/11

y = -4x/11 + 47/11

11y = -4x + 47

4x + 11y = 47

Let x be the units digit. The value of the hexadecimal integer in base 10 is  \(2 \cdot 16^3 + 15 \cdot 16^2 +1 \cdot 16 +x = 12048+x\)

The smallest multiple of 19 greater than 12048 is 12065. In order for 12048 + x to be a multiple of 19, x has to be 17 which is not possible in hexadecimal. Therefore, it is not possible.

NONE OF THEM WORK!

3x has to equal either 13, 25, and 37, but none of these nnumbers are divisble by 3. 

SO NO RESIDUES WORK :)

I graphed the answer, where I replaced t with x for the graph. I got

Thanks! :)

There actually is no minimum value. 

The larger x gets, the smaller the fraction becomes, therefore meaning there is no minimum value. 

The range is  \(y <1\), meaning no minimum value is present. 

Here's an image

The horizontal asymtote is \(y=0\)

Thanks! :)

If the number 5 is in the units digit, then the original number that was squared must end with 5. 

Let's set a variable to help us solve this question. 

Let's let a be a number that is the multiple of 10. We have

\((a + 5) (a + 5) = \\ a^2 + 10a + 25\)

a^2 will always end with a 0 or two 0s, so it has no effect on the final digit

10a will always end with a 0 or two 0s, so it has no effect on the final digit 

This means we always have 25. 

So 2 is our answer. 

Thanks! :)

I'm not clear on what the problem states, but I'm assuming it means that 0,1,2,3,4 are plugged in as x. 

So, see...

What the equation is saying is that x^5 must be divisble by 5. 

Let's plug in numbers

\(0^5 \equiv 0 \pmod{5} \) so it works

\(1^5 = 1 \equiv 1 \pmod5\) so it doesn't work

\(2^5 = 32 \equiv 2 \pmod5\) so it doesn't work

\(3^5=243\equiv3 \pmod 5\) so it doesn't work

\(4^5 = 1024 \equiv 4 \pmod 5\) so it doesn't work

So the only one that works is 0. 

Then again, residue is usually used as the remainder, but I hope my asusmption is correct.

Thanks! :)

First, let's convert 1/288 into base 13. 

Since 1 is the smae in every base, and we have

\((192)_{13} = (1 × 13^2) + (9 × 13^1) + (2 × 13^0) = (288)_{10}\)

This means that

\(\frac{1}{192}_{13} = \frac{1}{288}_{10}\)

Now, we complete base division. The answer is approximately 

\(0.0078229A44BC67118933...\)

I don't think the number is repeating, although I am not sure. 

If I had to geuss with the first 20 digits, it isn't repeating, but maybe it is. 

Thanks! :)

https://web2.0calc.com/questions/number-theory_13663

the answer is \(2\)

The answer is \(3\)

\(40 \cdot 6 \cdot 75 \cdot 12=216000\)

This number has 3 terminal zeroes. 

Thanks! :)

First, let's make a few observations. 

A power of 5 always ends with a 5. 

\(7^{13}=96889010407\), so 7^13 ends on a 7, 

And 23 ends with a 3. 

So we have \(7+5+3 = 15\). This means that \(5^{19} + 7^{13} + 23\) is divisible by 5.

Since 5, 7, and 23 are all odd, it's not possible the sum is divisble by 2. 

Also, since none of the terms are divisible by 3, the sum is not divisble by 3. 

Thus, 5 is our answer. 

Thanks! :)

Disclosure:  Maxematics gave me the idea how to find the answer       

when s/he pointed out that the product would be a multiple of nine,     

and therefore the digits in the product have to add up to a multiple     

of nine.  After that, I used brute force.    

_.    

A school orders $99$ textbooks, all for the same price.    

When the bill for the total order comes, the first and last    

digits are obscured. What are the missing digits?    
_18,486.7_    

What is the criterion for a correct answer?  If we can get       

a price per book that comes out even?  What about that      

the price of a book is a plausible amount?  You decide.       

                                               518,486.76     

                                              ––––––––––  =  5,237.24     

                                                      99              

_.