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8b^2c^5−4a^7b^3+10a^5c^4

Hint  :  add   the  exponents in  each  term

The  greatest of these  three  sums  is the  degree

If  you give me an  answer, I will check it......

f^-1 (50)=   7

f^-1  (10)  =  3

f^-1 ( 26)  = 5

So  we  have

f^-1   (  7 * 3  +  5)   =   f^-1 (26)  =   5 

                    x^2  - 6x + 8          x -  2

x =  0                   8                       -2

x = 1                    3                      -1

x = 2                    0                         0

x =  3                   -1                       1

x  = 4                   0                        2

x =  5                   3                        3

And  that's as  many solutions as  we  will  have  with  the  intersections of a parabola and a line

Angle AMD = 60 degrees.

The area of triangle AGN is 18.

Note  that

f(1)  =  floor  [ 1/ 6 ]  =  0

f(2)  = floor  [   0 ]  =    0

f(3) =   floor  [-1/8]  =  -1

f(4)  = floor  [ -2/ 9]  =  -1

.......

f(1000)  =  floor  [  -998 / 1005]  =  -1

So  the  sum  is    [  (1000 - 3   + 1  ] (-1)  =  [998] (-1)  =    -998

See my answer here  :

https://web2.0calc.com/questions/a-couple-of-questions

x + y + z = 4*sqrt(3).

The maximum volume is 600.

the answe is may 14

Since there are 30 days in april you add 1 week to 13 that is April 30. THen you add 2 weeks or 14 days. 

That is a good way too.

Plus it is reassuring that we got the same answer.  

To figure this out, I'll list the amount of money each type of change/bill adds up to.

Two dollar bill = 2.00

One dollar bill = 1.00

8 quarters = 2.00

11 dimes = 1.10

10 nickels = 0.50

3 pennies = 0.03

Now, I will add it all up.

2.00 + 1.00 + 2.00 + 1.10 + 0.50 + 0.03 = 6.63 dollars

To find the radius of a cone, you need not one, but two criteria to derive it. Because all that is given is the volume, it is not possible to find the radius of this cone unless given the height, too. This is because to determine a variable such as the radius, we must have all other criteria for the formula to find the volume to figure it out. Because the formula for the volume of a cone is 1/3 * pi * r^2 * h, we need the height to find out the radius.

So it is not possible.

But, if you were to have said a sphere, we would be able to figure it out because the volume of a sphere is 4/3 * pi * r^3 so there are no other criteria except for the radius needed to figure out the volume.

Now, I noticed that the question title actually says "volume of a pyramid", so maybe there was a typo. Still, this would be impossible due to the fact that the volume of a pyramid formula states, \({\displaystyle V={\frac {1}{3}}lwh},\) where l is the slant height, w is the width, and h is the height.

AD, 1 of each letter, 5! ways to arrange: 120

AD, 2 of one letter, 1 of another, 6 possible choosing of letters, 60 ways to arrange: 360

AD, 3 of one letter, 3 possible ways to choose, 20 ways to arrange: 60

60+360+120

Total 540 ways.  :))

=^._.^=

Ohhh, that's smart. 

=^._.^=

Since there are no restrictions on the strings, we can think of this problem as drawing 5 random digits in sequence. There is a 9/10 chance each time we draw to NOT get a 4. Every time we draw, we multiply by 9/10 again until we've draw 5 digits. Thus, the answer is (9/10)^5 :)

I would say each pencil costs $2.

We would set up an equation:

$24 = 7x + $10 where x is the price per pencil.

Now, we would substitute values in for x:

$24 = 7(1) + $10

$24 = 7 + $10

$24 = $17 <- Not True

$24 = 7(2) + $10

$24 = 14 + $10

$24 = $24 <- True!

So, each pencil costs $2.

Hope this helps. :)

I would say you could purchase 3 balloons.

If each balloon costs $2 and you need to pay a $3 delivery fee, then you would set up an equation:

p = $2x + $3 where p is the total price and x is the number of balloons.

Now we would substitute different numbers in for x:

p = $2(1) + $3

p = $2 + $3

p = $5

p = $2(2) + $3

p = $4 + $3

p = $7

p = $2(3) + $3

p = $6 + $3

p = $9

Now, knowing that we have $9 to spend, the max amount of balloons we can buy is 3. :)

do the substitution and work it out.

x=1    and    f(1)=1

We must have exactly 1 A and exactly 1 B  the other 3 letters can be C,D,E repeats allowed, so that is 3^3=27 ways with order counting.

The other A and B must be slotted into the string.        5C2=10 places but either the A or B can go first so that is 20 places.

So I get 27*20 = 540 ways.

For how many positive integers n > 1 is it true that 2^{24}*2^{36} is a perfect nth power?

2^60, (2^2)^30, (2^3)^20, (2^4)^15, (2^5)^12, (2^6)^10, (2^10)^6, (2^12)^5, (2^15)^4. (2^20)^3, (2^30)^2

You can count.

Answered on the earlier version.

Next time please refer people to the earlier post in your question.

Thanks for the address guest:

Let the amount he has to start with be x where x >=200

She puts $50 in her left pocket leaving x-50 remaining.

She gives away 2/3(x-50) leaving 1/3(x-50) to go into her right pocket.

So now she has  50+1/3(x-50)

We are told that this is more than x-200

and you can finish it.

∠DAC + ∠AEF = 180º

∠DAB = ∠BAC 

∠BAC = (∠DAC + ∠AEF + 12) / 12

∠BAC = 16º

∠DAC = 32º