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3 years ago

Answer Question Below

Mathematics

2 answers:

MAVERICK [17]3 years ago

7 0

Ima guess with 1/3. i’m not sure

damaskus [11]3 years ago

3 0

It’s 1/3 it’s not a negative that day

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a long board 4.7 thick is cut into three pieces of equal length these three pieces are then stacked the stack is cut in three eq

nalin [4]

The number of pieces per stack after 3 cuts represented by a power expression can be written as $3^{3}$

The number of equal pieces per cut = 3

The number of cuts made = 3

Using a power expression :

The number of of equal pieces per cut is raised to a power which represents the number of cuts made

Thus we have $number\: of \:equal\: pieces^{number\: of \:cuts}$

After 1 cut ; $3^{1} = 3$
After 2 cuts ; $3^{2} = 3\times3 = 9$
After 3 cuts ; $3^{3} = 3\times3\times3 = 27$

Therefore, the Number of pieces present in stack after 3 cuts is $3^{3}$

Learn more :brainly.com/question/18796573

5 0

3 years ago

Y=2x+1 How do I put this equation in order pair for a table

Ber [7]

Answer:

I think you would have to draw lines like is the graph book, you will see that they have two axes which is the y and the x.You will then have to fix in numbers and indicate the y= 2x+1.

Step-by-step explanation:

hope it helps

8 0

3 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$