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yea, sorry, I accidentally typed it wrong last time

I don't know if the numbers in the question have been changed since MathWizPro answered, but I see the following question:

"The sum of two numbers is 3 more than four times the first number. Their difference is 10 less than twice the second number. Find each of the numbers."

Oops srry nvm 

This is the concept of algebra, given that ab^4=12 and a^5b^5=7776, the value if a will be found as follows:
ab^4=12
a=12/b^4
also;
a^5=7776/b^5
thus;
a=(7776/b^5)^(1/5)
a=6/b
thus the value of a will be:
6/b=12/b^4
dividing both sides by b we get:
6=12/b^3
multiplying both sides by b^3 we get
6b^3=12
b^3=2
hence;
b=2^(1/3)

I'm pretty surre I already answered this...

This is the concept of algebra, given that ab^4=12 and a^5b^5=7776, the value if a will be found as follows:
ab^4=12
a=12/b^4
also;
a^5=7776/b^5
thus;
a=(7776/b^5)^(1/5)
a=6/b
thus the value of a will be:
6/b=12/b^4
dividing both sides by b we get:
6=12/b^3
multiplying both sides by b^3 we get
6b^3=12
b^3=2
hence;
b=2^(1/3)

I hope that was helpful, and also, welcome!

Simplify the following:
sqrt(8) + 3 sqrt(2) + sqrt(32)

sqrt(8) = sqrt(2^3) = 2 sqrt(2):
2 sqrt(2) + 3 sqrt(2) + sqrt(32)

sqrt(32) = sqrt(2^5) = 2^2 sqrt(2):
2 sqrt(2) + 3 sqrt(2) + 2^2 sqrt(2)

2^2 = 4:
2 sqrt(2) + 3 sqrt(2) + 4 sqrt(2)

Add like terms. 2 sqrt(2) + 3 sqrt(2) + 4 sqrt(2) = 9 sqrt(2):

9 sqrt(2)

I rarely if ever answer questions that have not been asked properly, all those $ signs have no meaning.

Plus you will have a slightly higher chance of getting good answers if you become a member. (it is free and easy) 

If you make a good impression as a member it becomes almost a given that you will get a good answer.

Answered.

\(\overline{FG}=\sqrt{3.5^2+7^2}\\ \overline{FGE}=2\sqrt{3.5^2+7^2}\\ [\overline{FGE}]^2=4(3.5^2+7^2)\\ [\overline{FGE}]^2=245\\ l^2=245 \)

Here is my attempt at this:
Winter growth = W
Summer growth =4W
Fall growth =F
Spring growth =2F

Spring shedding =F
Fall shedding =11F
Winter shedding =W
Summer shedding =1/12W
5W + 3F =1000,
13/12W + 12F =855, solve for F, W
F =~56 Leaves
W =~166 Leaves
Total Spring and Fall growth =56 * 3 =~170 leaves (rounded). 

There are 27 ways of arranging the marbles.

If there were 5 different letters, there would be 5! ways of arranging them.

However, since there were two copies of the letter X, we need to divide that answer by 2!.

And, since there were two copies of the letter E, we need to divide that answer by 2!.

5! / 2! / 2!  =  120 / 2 / 2  =  30 different arrangements of the letters.

  Frequently happens to me too ! 

Hi thank u ! I had a major brain f**t and couldn't wrap my head around it for some readon :-)

$12 represents  9 % of  $ x              where x = original price

9% * x = 12

.09x=12

x = 12/.09 = $ 133                   then the price increases 12 dollars  = 145 dollars price

xº = (80º -12) / 2

As follows:

\(\dfrac15 \div 4 = \dfrac15 \div \dfrac41 = \dfrac15 \times \dfrac14 = \dfrac{1\times 1}{5\times 4} = \dfrac1{20}\)

Everything outside on the surface of the cube is visible.

That means the taking away the outside layer, the inner 8 x 8 x 8 cube would not be visible.

Total number of visible 1 x 1 x 1 cubes = \(10\times 10 \times 10 - 8 \times 8 \times 8 = \boxed{488}\)

First, we find \(f^{-1}(x)\) in terms of a and b.

\(f(x) = ax + b\\ x = af^{-1}(x) + b\\ f^{-1}(x) = \dfrac{x - b}a\)

Now,

\(g(x) = f^{-1}(x) - 3\\ 5x - 4 = \dfrac xa - \left(\dfrac ba + 3\right)\)

Comparing coefficients:

\(\dfrac1a = 5\\ a = \dfrac15\)

\(5b + 3 = 4\\ b = \dfrac15\)

Therefore

\(5a + 5b = 5\cdot \dfrac15 + 5\cdot \dfrac15 = 2\)

Adding the 5 equations:

4(a + b + c + d + e) = 84

a + b + c + d + e = 21 --- (*)

Subtracting each equation from (*),

e = 21 - 14 = 7

d = 21 - 15 = 6

c = 21 - 17 = 4

b = 21 - 18 = 3

a = 21 - 20 = 1

abcde = 1(3)(4)(6)(7) = 504. 

Using the definition:

\(\cos B = \dfrac6{10} = \dfrac{AB}{BC}\)

From this, we can conclude AB = 6.

By Pythagorean theorem, AC = 8.

\(\tan C = \dfrac{AB}{AC} = \dfrac68 = \boxed{\dfrac34}\)

If the sum of the first n terms of an arithmetic progression is \(\dfrac{3n^2+13n}{2}\),
find the 25th term of this progression.

\(\begin{array}{|rcll|} \hline \mathbf{s_n} &=& \mathbf{\dfrac{3n^2+13n}{2}} \\\\ s_{25} &=& s_{24} + a_{25} \\\\ a_{25} &=& s_{25}- s_{24} \\\\ a_{25} &=& \dfrac{3*25^2+13*25}{2}- \dfrac{3*24^2+13*24}{2} \\\\ a_{25} &=& 1100 - 1020 \\ \mathbf{a_{25}} &=& \mathbf{80} \\ \hline \end{array}\)

The 25th term of this progression is \(\mathbf{80}\)