All Answers 358108 Answers

31753  =  113 x 281

Answer:

\([200, 350)\)

Explanation:

Let \(x\) be the number of dollars in the pile of money. Since she puts \($50\) in her left pocket, gives away \(\frac{2}{3}\), of the rest of the pile, and then puts the rest in her right pocket, Jeri has \(50+\frac{1}{3}(x-50)\) dollars. 

We also know that she would have more money than if instead she gave away \($200\) and kept the rest. With this information, we can write an inequality. 

\(50+\frac{1}{3}(x-50)>x-200\) .

Multiplying each side by \(3\), gives us 

\(150+(x-50)>3x-600\).

When we group like terms we get

\(x+100>3x-600\).

Then, we subtract \(100\) from both sides, giving us 

\(x>3x-700\).

Subtracting \(3x\) from each side gives us

\(-2x>-700\).

When we multiply each side by \(-1\), this also reverses the inequality so, we have

\(2x<700\).

Dividing each side by \(2\), gives us

\(x<350\).

 Since we know \(x\ge200\) but \(x<350\), we can write the interval \([200, 350)\).

\(\)

Let d be the number of dollars that Jeri finds.  Then solving the inequalities, you get d >= 200 and d < 575, so the answer is [200,575).

Thank you all so much for the helpful answers!!!!

Here is the solution given:

Let \(q(x) = (x^2 - 1) p(x) - x.\) Then \(q(n)\) has degree 7, and \(q(n) = 0\) for n=2, 3, 4, ... 7, so \(q(x) = (ax + b)(x - 2)(x - 3) \dotsm (x - 7)\)
for some constants \(a\) and \(b.\)

We know that \(q(1) = (1^2 - 1)p(1) - 1 = -1.\) Setting \(x=1 \) in the equation above, we get \(q(1) = 720(a + b),\)
so \(a+b=-\frac{1}{720}.\)

We also know that \(q(-1) = ((-1)^2 - 1)p(-1) + 1 = 1.\) Setting \(x=-1\) in the equation above, we get \(q(-1) = 20160(-a + b),\)
so \(-a + b = \frac{1}{20160}.\) Solving for \(a\) and \(b,\) we find \(a = -\frac{29}{40320}\) and \(b = -\frac{3}{4480}.\) Hence, \(\begin{align*} q(x) &= \left( -\frac{29}{40320} x - \frac{3}{4480} \right) (x - 2)(x - 3) \dotsm (x - 7) \\ &= -\frac{(29x + 27)(x - 2)(x - 3) \dotsm (x - 7)}{40320}. \end{align*}\)
In particular, \(q(8) = -\frac{(29 \cdot 8 + 27)(6)(5) \dotsm (1)}{40320} = -\frac{37}{8}, \)
so \(p(8) = \frac{q(8) + 8}{8^2 - 1} = \boxed{\frac{3}{56}}.\)
 

The answer is 6.

tan C = 3/5.

Will has read 20 * 5 = 100 pages 

that is, Will had read 100 pages

But it is given that, after reading for 5 days, Will still had 55 pages left

therefore 20 * 5

               = 100

               = 100 + 55 = 155

Total number of pages in the book = 155

By Pythagoras, [EFGH] = 40.

I dont get what you are saying and that is incorrect answer

They can overlap by 4 cm^2, so the minimum surface area is 6*2^2 + 6*3^2 - 4 = 74 cm^2.

130th digit is 6

Here is Alan's answer graphed, from Desmos.

I wouldn't guess that this was a degree 5 polynomial.  (without doing lots of work)  

Although it would have to be an odd degree.  That makes 5 reasonable I guess.

Thanks Alan.    

The positive integers from 1 to 150 inclusive are placed in a 10 by 15 grid so that each cell contains exactly one integer. Then the multiples of 3 are given a red mark, the multiples of 5 are given a blue mark, and the multiples of 7 are given a green mark. How many cells have more than 0 marks?

Here is how to start it.

Draw a Venn diagram

The bit that overlaps the most has the number of multiples of all of them in it. 

3*5*7=105   There is only one multiple of 105 in that set of numbers.

Now work out from there.

How many multiples of 3 and 5 are there?   3*5=15       150/15=10      10-1=9

So where are you going to put the 9?

Yes, the polynomial is:

p(x) = (-29/40320)x⁵ + (3/160)x⁴ + (-1571/8064)x³ + (163/160)x² + (-3081/1120)x + (27/8)

Thanks Alan,

 So that means there is a polynomial of degree 5 that fits?  

Good one Ginger :)

This is the result I get, using Mathcad:

No; aces only becomes Faces by adding an “F”

Solution:

\(\text {52-card standard deck. }\\ \text {12 face cards}\\\)

\(\text {There are } \binom{52}{5}\text { ways to draw five cards.}\\ \text {There are }\binom{12}{5} \text { ways to draw five face cards.}\\ \, \Large P_{\small 5 \text { face cards}} = \dfrac {\binom{12}{5}}{\binom{52}{5}}= \dfrac {33}{108290} \approx 0.03047\%\)

GA

--. .-

The function f(x) = ax^r satisfies f(2) = 1 and f(32) = 16. Find r.

\(1=a*(2^r) \qquad 16=a*(32^r)\\ 1=a*(2^r) \qquad 2^4=a*(2^{5r})\\ 2{-r}=a \qquad \qquad 2^{4-5r}=a\\ 2-r=4-5r\\ etc\)

LaTex:

1=a*(2^r) \qquad 16=a*(32^r)\\
1=a*(2^r) \qquad 2^4=a*(2^{5r})\\
2{-r}=a \qquad \qquad 2^{4-5r}=a\\
2-r=4-5r\\
etc

She has 12 moves.   

So that is 2^12 possible paths

Let her take r moves clockwise  (positive)  and 12-r moves anti-clockwise.  (negative)

So she will end up at   r - (12-r) = 2r-12 in a positive directions which is

2r-12 (mod6) = 2r (mod6) 

2r(mod6) = 0 when r = 0,3,6,9,12

If r=0 then she goes anticlockwise every time so there is only one way she can do that      1

If r=12 then she goes clockwise every time so there is only one way she can do that          1

If r= 3 there are  12C3 = 220 possible paths

If r = 9  there are also   220 possible paths

If r=6 there are 12C6 = 924 possible paths

924+440+2 =  1366 possible paths where she will end up back at A

Probability that she will end up back at A

\(=\frac{1366}{2^{12}}\\ =\frac{683}{2^{11}}\\ \approx 0.335 \qquad\text{Correct to 3 decimal places}\)

**    I am not claiming to be positive that this is completely correct.

Thanks Ginger.

What calculator/program  did you use to get those values Gino? 

See post #11, below.