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3 years ago

7

checkers is one of the oldest board games still played today. the oldest known checkerboard is over 5,000 years old. a standard

checkerboard is square and has an area of 64 square inches. what is the length of one side of the board? PLEASE HELP

1 answer:

Sav [38]3 years ago

4 0

Answer:

I think its 8 because 8 plus 8 is 64

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A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

7 0

4 years ago

Graph the Rarabola.---Plot five points on the parabola: the vertex, two points to the left of the vertex, and tvbutton2Explanati

valentinak56 [21]

Which parabola?

$y=-x^2$

Equation of the parabola

y - y1 = 4p(x - x1)

The Vertex of the parabola given is (0, 0) because it does not have the values of x1 and y1.

Then Vertex = (0,0)

-Look for two values of y to the left and two points to the right.

You can choose the points that you which

x y

-3 y = -(-3)^2 = -9

-1 y = -(-1)^2 = -1

0 y = -(0)^2 = 0

1 y = -(1)^2 = -1

3 y = -(3)^2 = -9

3 0

1 year ago

Three times a number increased by four is<br> less than -62.

klemol [59]

Answer: x<-22

Step-by-step explanation:

3x+4<-62

3x<-66

x<-22

6 0

3 years ago

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Find the value of x if ∡A and ∡B are supplementary angles.

Elanso [62]

Answer:

$x = 30.5$

Step-by-step explanation:

Two angles are supplementary when they add up to 180°.

So we know that:

$C + D = 180$

Supplanting we get:

$(3x-44) + (5x-20) = 180$

We add up the X's.

$8x-44-20 = 180$

$8x-64 = 180$

Then we send 64 to the other side adding:

$8x = 180+64$

$8x = 180+64$

$8x = 244$

And then the 8 goes dividing:

$x = 244 : 8$

$x = 30.5$

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3 years ago

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Rufina [12.5K]

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