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Nor do I!

Because the number must be a multiple of 5, the final digit must be a 5. 

There are \(5! = 120\) ways to order the other digits. 

For the number to \(>450,000\), the numbers can be in the order \(46\text{_,__}5\) or \(6 \text{__,__5}\)

For the first case, there are \(3! = 6\) ways to order the 3 remaining digits. 

For the second case, there are \(4! = 24\) ways to order the remaining 4 digits. 

Thus, the probability is \({24 + 6 \over 120} = \color{brown}\boxed{1 \over 4}\)

Don't see anything !!!

Thank You!

1.

Either 

5x^2  =   -7        impossible

or

21 + x  = -7    ⇒  x  =  -7 -21  =  -28   but this is not in the  domain of the  second function

or

2 - sqrt (x)    =  -7

So

sqrt (x) =  2 +7

sqrt (x)   =9        square both sides

x   =81              this is in the domain of the  third function

So   g^-1(-7)  =   81

Let the price of the toaster==T
The price of the kettle ==2/5T

T + 2/5T ==105, solve for T
T ==$75 - price of the toaster after 20%discount.
75 x 2/5==$30 -price of the kettle before 20% discount.

T x 0.80 ==$75, solve for T
T ==75 / 0.8 ==$93.75 - price of the toaster before 20% discount
133 - 93.75 ==$39.25 - price of the kettle before 20% discount.

1 - [30 / 39.25] ==23.567% - discount percentage on the kettle. 

In a sale Tommy bought a kettle and a toaster. He paid a total of $105. The amount paid for the kettle is 2/5 the cost of the toaster. The total price before the discount should have been $133. The toaster was reduced by 20%. How much was the kettle reduced by?

Call the original prices K , T

K + T =   105

(2/5)T + T  =105

(7/5)T  = 105

T = 105 * 5   / 7  =    $75

So

K  = 133 -  75  =  $58

But with the discount the toaster now costs  75 (.80)  = $60

So  ......the kettle's new cost must be  105 -  60 =   $45

So.....the discount on the  toaster  =   1  - 45  / 58   ≈ .224 ≈  22.4%

Let the amount Paid for toaster = x

the amount Paid for Kettle = 2x/5

x + 2x/5 = 105

or (5x - 2x)/5 = 105

7x = 525

x = 75

Cost of Toaster = $75

Cost of Kettle = 2/5 * 75 = $30

Discount on Toaster = 20%
actual discount on toaster 

= 20/100 * 75 = $15

Total discount = 133 - 105 = $28

discount on Kettle = 28 - 15 = $13

Kettle was reduced by $13

No, but I found the answer already by guessing and checking, \(x = 16\), \(y = 15\) and \(z = 3\)

That's interesting, I wonder why it works....

I found online that you can do

 \(720≥z^3 \\ 1≤𝑧≤8 \\ \),

Then you try out 8 cases

I did that, and I just brute forced it from here...

Surely there's a better way...

2.5 L =   2500 cm^3

So

2500  =   pi * r^2  * height

2500 = 17 pi  * r^2

r^2   =   [ 2500 / (17/pi ] 

r = sqrt [ 2500 / (17 pi ) ]  ≈  6.84 cm

A discount of 25%  means  the price is 75%  of what it originally cost....and this  is  taxed at  12%

So....

cost =     (186)(.75) ( 1.12)   =  $156.24

did u just answer ur own question

if so.. nice job!

The small cylinder can fill the big cylinder to a height of 5 cm

work out the radius of the big cylinder

small cylinder radius=6cm

small cylinder height=12 cm

Problem 1) PQ is similar to AB because of Side-Angle-Side. P and Q lie on sides AO and OB respectively, and \(\angle POQ = \angle AOB\).

Problem 2) Because \(OP = {3 \over 5} \times AO\), we know that \(OP = {3 \over 5} \times 10 = \color{brown}\boxed{6}\)

Because \(PN = 8 \), \(QN = 10\), and \(\angle N = 90\), \(PQ = \sqrt{164} = 2\sqrt{41}\)

We know that \(M\) is the midpoint, so \(PM = \sqrt{41}\).

Now, draw \(MA\), so that \(MA \) is perpendicular to \(PR\).

Because of similar triangles, \(MA = 5\) and \(AN = 4\)

Using the Pythagorean Theorem, we find that \(RM = 13\).

Because of similar triangles, \(OR = {2 \over 3} \times 13 = \color{brown}\boxed{26\over3}\) 

In triangle PMR  Draw atltitude MS   to  PR

Now  triangles  MPS and  QPN are similar

So    MP  /  QP   =  PS  / PN       

          1/2 =  PS /PN   ⇒   PS   = 4   ⇒  MS = (1/2) QN  =  5

And triangles   RSM  and  RNO   are similar

So    SR   = 12      MS  = 5     RN  =  8

MS / SR   =  ON  / RN

5 / 12  =  ON / 8

40 /12  =  ON   =   10/3

And since   triangle RNO  is right, then

OR   = sqrt   [ RN^2  + ON^2  ]   =   sqrt  [  8^2  + (10/3)^2 ]  =

sqrt [  64 +  100 / 9  ]  =  sqrt  [ (576 + 100) / 9  ]   =   sqrt  [ 676 ] / 3  =   26  / 3

We assume \(p(x) = x^4 + ax^3 + bx^2 + cx + d\).

Since p(1) = 3, \(a + b + c + d = 2\).

Since p(3) = 11, \(27a + 9b + 3c + d = -70\).

Since p(5) = 29, \(125a + 25b + 5c + d = -596\).

This system has 3 equations but 4 variables, so there are infinitely many solutions.

Solving, we have \(\begin{cases}b = -9a - \dfrac{227}4\\c = 23a + 191\\d = -15a - \dfrac{529}4\end{cases}\).

With these coefficients, we have \(p(-2) = 16 - 8a + 4\left(-9a - \dfrac{227}4\right) - 2(23a + 191) + \left(-15a - \dfrac{529}4\right) = -\dfrac{2901}4 - 105 a\)

Also, \(p(6) = 1296 + 216a + 36\left(-9a - \dfrac{227}4\right) + 6(23a + 191) + \left(-15a -\dfrac{529}4\right) = \dfrac{1067}4 + 15 a\).

Hence, 

\(\begin{array}{rcl}p(-2) + 7p(6) &=& -\dfrac{2901}4 - 105a + 7\left(15a + \dfrac{1067}4\right) \\ &=&1142\end{array}\)