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 x  is the second number     and    x- 4  is the first number

4 ( x-4)  -14 = 5 x

4 x - 30 = 5x

x = - 30      x-4 = -34

The imaginary part of a is 3 - sqrt(3).

sqrt (  sqrt5 / sqrt (625*5))

sqrt ( sqrt 5 / (25 sqrt 5) )

sqrt (  1/25)

1/5

The width of the interval is 12.

Using exponential form, the imaginary part of z/w is sqrt(2)/2.

The product of all possible values of c is 350.

update: i got the answer it was 7/4. thanks for the help tho!!!

i think you messed up when you added
 

-2x - 2y - 6z  =  -20

-4x + 3y + 5z  =  7

because -2y+3y doesnt cancel out 🙂

let me try using your way of doing to see if i can get a correct answer

sorry for the bother but it isnt correct. i need some help with this

Like so:

Like so:

You are welcome :)

Thank you very much!!!
Yes, the book assumes \(g=10ms^{-2}\).

In order to get a to be 1.04ms^-2 you have to assume a gravitational acceeration g = 10ms^-2 (rather than the more normal 9.8 or 9.81).

What is the equation?

You mean 9² 

[ABCD] + [AEFG] = 81 square units

Let's see

\(f(x)=\frac{x^5-1}3, Find f^{-1}(-31/96) \)

\(f(x)=\frac{x^5-1}{3}\\ y=\frac{x^5-1}{3}\\ 3y=x^5-1\\ 3y+1=x^5\\ x=\sqrt[5]{3y+1}\\~\\ f^{-1}(x)=\sqrt[5]{3x+1} \)

\(f^{-1}(-31/96)\\ =\sqrt[5]{3*\frac{-31}{96}+1}\\ =\sqrt[5]{\frac{-31}{32}+1}\\ =\sqrt[5]{\frac{1}{32}}\\ =0.5\)

LaTex:

f(x)=\frac{x^5-1}{3}\\
y=\frac{x^5-1}{3}\\
3y=x^5-1\\
3y+1=x^5\\
x=\sqrt[5]{3y+1}\\~\\
f^{-1}(x)=\sqrt[5]{3x+1}

Because both circles are tangent to the same spot, the centers of the circles and the point of tangency form a line. There are two distances: one where both circles are on the same side, or where the circles are on seperate sides. 

If the two circles are on the same side, the distance would be 8 - 3 = $\boxed5$

If the two circles are on different sides, the distance would be 8 + 3 = $\boxed{11}$

The median of a trapezoid's length is the median of the two bases. So the median of ABCD is 12. Similarly, the median of ABMN's length is the median of the two bases AB and MN, so the length of PQ is $\boxed{8\frac12}$

If a and b must be positive integers, then there can be 29 ordered pairs :

1 + 29

2 + 28

...

28+2

29+1

There won't be 30 because 0 is not concidered positive or negative.