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Each X represents where one block would be. The other possible combinations would involve transforming the L \(1-3\) units to the right.

Woah.... that is really smart!

I have not thought about it that but what do you means the bottom 2 pieces are in the bottom row

There are \(4\) ways to arrange the L so the bottom \(2\) peices are in the bottom row.

There are \(3\) rows that you can do this on, so there are \(12\) ways to arrange the L this way. 

There are \(3\) additional ways to rotate them \(90^\circ\) with the same concept. This means that there are a total of \(48\) ways with rotations.

There are \(3\) ways to reflect the L , with each one having \(12\) ways to arrange them. 

This makes for a total of \(\color{brown}\boxed {84}\) ways.

Ye like this

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the dots mean 1 block

The 4 squares are unit squares that form an L shape that is 3 high and 2 wide right?

There are \(36\) possible outcomes (\(6\) for the first roll and \(6\) for the second roll).

The only square numbers in the list are \(1\) and \(4\).

If we don't want either of these, there is \(4\) sucsessful options for the first role and \(4\) sucsessful options for the second roll. This means that there is \(16\) succsessful outcomes, out of the \(36\) total outcomes, meaning that the answer is \(16\over 36\), which can be simplified to \(\color{brown}\boxed{4\over9}\)

For any positive integer to satisfy the conditions, it has to be a perfect power of \(2\), and it should be able to be equivalent to \(2^{42}\), with integer values.

The only possible options are:

\(2\), can be written as \(2^{42}\)

\(4\), can be written as \(({2^2)}^{21}\)

\(8\), can be written as \((2^3)^{14}\)

Thus... the answer is \(\color{brown}\boxed3\)

That's what we're here for!

The possible pairs of numbers are \((0,3), (1,4), (2,5), (3,6) ,(4, 7),(5,8),(6,9)\)

Each of the \(7\) pairs can be ordered in \(2\) ways, making for \(14 \) numbers.

However, we need to account for the fact that the number \(03\) is not a \(2\)-digit number, so we have to subtract \(1\), making our total \(\color{brown}\boxed{13}\)

Thanks that helped so much

We can write this problem as an equation: \(11 \times 8.5 \times h = 140.25\).

In this equation, \(h\) stands for the height of the \(400 \) papers. 

By solving for \(h\), we find that the height is \(1.5\) inches,

However, we are not done yet...

We need to account fot the fact that \(h\) is the height of \(400\) sheets of paper.

This means that the height for each paper is \(1.5\over {400}\), which can be simplified to \(\color{brown}\boxed {3\over{800}}\)

A carry occurs when the place values have a sum \(\geq10\). 

For the \(1\)'s digit, there are \(7 \) choices \((0,1,2,3,4,5,6)\)

For the \(10\)'s digit, there are \(4\) choices \((0,1,2,3)\)

For the \(100\)'s digit, there are \(5\) choices \((1,2,3,4,5)\). Remember, we can't use \(0\) for the hundred's digit, because then the number wouldn't be 3-digits.

This means that there are \(7 \times 4 \times 5\) numbers. Thus, the answer is \(\color{brown}\boxed {140}\)

I'm going to assume for this problem that the zarks are the zacks. If they are different, this problem can't be solved.

7 zacks = 1 zelk

14 zacks = 2 zelks

2 zacks = 6 zinks

14 zacks = 42 zinks

Thus... the answer is \(\color{brown}\boxed{42\space\text zinks}\)

Hector     20     22     24

partner    11     22                          on the second set they will be lifthing the same amount  22 #

They are both the same length

Using Pythag Theorem

s^2 + s^2 = 1^2

s = sqrt ( 2)  / 2

699/900 = 233/300

hints: 

AE: sin 45 degrees = AE/1

CE: cos 45 degrees = CE/1

using the fact that:

sin = opposite/hypotenuse

cos = adjacent/hypotenuse

easy way to remember sin, cos, & tan: sohcahtoa

hope this helped.

-c

right, thank u! :]

For starters, \(y=0\), because then \(x=0\).

However, there is more than \(1\) solution. 

Here is the work in LaTex:

\(x+xy=250y\)

\(x=250y-xy\)

\(x=(250-x)y\)

\({x\over{250-x}} = y\)

\(x- {{x^2}\over{250-x}}=-240{x\over{250-x}}\)

\(x- {{x^2}\over{250-x}}={-240x\over{250-x}}\)

\(x={{-240x+x^2}\over250-x}\)

\(250x-x^2=-240x+x^2\) 

\(490x=2x^2\)

\(245x=x^2\)

\(x=245\) 

Plug it in to the equation: 

\(245+245y=250y\)

\(245 = 5y\) 

\(y=49\)

This means that the two answers are\(\color{brown}\boxed {y=0,y=49}\)

Simplify the first equation to \(4x=-8-ay\).

Simplify the second equation to \(3x=b+y\).

Multiply this equation by \(4 \over 3\).

This equals \(4x= {4 \over 3} b + {4\over3}y\).

For \(4x=-8-ay \) and \(4x={4\over3}b+{4\over3}y\) to have infinite intersection points, they have to be the same. 

So... \(\color{brown}\boxed {a=-{3\over4}, b = -6}\)

Need the graph to answer first and third parts.....

It is actually \(37^ \circ\) Guest. This is \(\angle GCE = \angle CBJ\). \(\angle DBF \) is opposite to \(\angle CBJ\), which means that \(\angle DBF = \angle CBJ\). This means that \(\color{brown} \boxed {\angle ECG = 37^\circ}\)

Triangle side rule says the third side must be less than the sum of the other two sides

33 + 42 = 75     so side three must be < 75     side three = 74  ( integer length)   

74   + 33 + 42 = peimeter