proyaop

Order of operations:

First do parenthesis

Then do exponents

Then do multiplication (and if there is a negative sign treat it as -1 * (_))

Then do division

Then do addition

Then do subtraction

Abbreviation being PEMDAS... good luck...

Because each variable has a positive or negative value, we have to account for both possible values, creating two "cases", or options, and we test each one to get a different solution/ordered pair. If you don't understand, you should try searching "casework" problems, but this problem is just getting used to algebraic expansion. Remember difference of cubes!!!

(b)

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Earlier, we manipulated a^2 + ab + b^2 into (a - b)^2 + 3ab, however you can also manipulate it to (a + b)^2 - ab (do the expansion and see that it works!)

So... a^3 - b^3 = (a - b)[(a + b)^2 - ab] = 52

However, from (a), we learned that a - b = 4, and that ab = -1, substituting in, we have a^3 - b^3 = 4[(a + b)^2 + 1] = 52.

Dividing by 4, we have (a + b)^2 + 1 = 13, a + b = positive/negative square root of (12), which sqrt(12) = 2sqrt(3), so positive negative 2sqrt(3).

Hence, \(a + b = \pm2\sqrt{3}\)

You can use distributive property => f(x) = -12x + 4

Domain is legal inputs for x, so domain is "all real numbers"

Range is legal outputs of f(x) (or values of f(x)), so range is "all real numbers"

\(x = {1\over2 + {1\over x - 3}}\), multiply by (x - 3)/(x - 3)

\(x = {x - 3\over 2x - 6 + 1}\)

2x^2 - 5x = x - 3

2x^2 - 6x + 3 = 0

Quadratic formula!!!

\((x - {1\over x})^2 = x^2 - 2 + {1\over x^2} = 3 - 2 = 1\)

\(x - {1\over x} = \pm1\)

Please post your question if you need help... LOL

1.4268292682926829^(1/20) or that^0.05 = 1.0179316092382140458823473304434

Scientific calculators go down to the last decimal... No need to post this question because this website has a literal calculator :O

Yes there definitely is a good strategy. We have 10 different variables.

Case 1, the sum is = 0:

There is 1 solution, where all variables are equal to 0.

Case 2, the sum is = 1:

There are 10 choose 1 ways to arrange the value of 1 among the 10 variables, so there are 10 solutions.

Case 3, the sum is = 2.

We can have 10 ways to arrange the 2. But if we have two 1s, we have (10 choose 2) ways to do it = 45.

Total sum is 10 + 45 + 10 + 1 = 66 total cases.

Guest, please stop posting random answers. 

We know PQ is 18, and by pythagorean theorem, we can conclude that RS (the points of tangency of circles P and Q respectively) has length sqrt(18^2 - 2^2). sqrt(320) simplifies to 8sqrt(5). (unnessecary but you can do it this way)

Let QR = x. By similar triangles, x/8 = (x + PQ)/10

10x = 8x + 8*18

2x = 144

x = 72 = QR

There is no diagram below, please repost your question with "the diagram below" please, thank you.

8x^3 = sqrt(128) = 8sqrt(2)

x^3 = sqrt(2) = (2)^0.5

So x = (2)^(1/2)(1/3). Thus, n = 6.

BDI is not a triangle by the triangle inequality, please talk to your teacher/problem owner to tell them of this error/typo.

Remember: (You can simplify by expanding out)

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

and

a^2 + b^2 = (a + b)^2 - 2ab

Substituting in, you get:

(a + b)(a^2 - ab + b^2) = 812 + (a + b)^2 - 2ab

14[(a + b)^2 - 2ab - ab] = 812 + 196 - 2ab

14(196 - 3ab) = 1008 - 2ab 

2744 - 42ab = 1008 - 2ab

1736 = 40ab

ab = 43.4

a + b = 14

Write a quadratic equation using vieta's formula:

x^2 - 14x + 43.4 = 0. x is the values of a and b.

a and b are 14 +- sqrt(22.4)) / 2

They have a common denominator so you can multiply both sides by a^3 + 7.

You obtain the equation 1 = -a^3, a = -1

Combine like terms, so combine the ones with the same variable or the ones with the same constant/normal numbers and you get:

5c + d + 1