Ask question

Ask question

All categories

2 years ago

12

a checheers board is 8 square long and 7 squares wide. The area of each square is 14 square centimeter. Estimate the perimeter o

f the checkers board to the nearest tenth of a centimeter

1 answer:

Lesechka [4]2 years ago

4 0

Turn 14 into 16, 4 x 4 = 16, 4 x 8 = 32, 32 x 4 = 128cm

You might be interested in

Consider the function, f(x)=2x-1

sukhopar [10]

Answer:

1, -1 , -3 ,3

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

IS THIS GROWTH OR DECAY

Maksim231197 [3]

Answer:

It is a growth.

Step-by-step explanation:

If the number in parenthesis (exponential base) is greater than 1, then it is a growth. If it is between 1 and 0, then it is a decay.

4 0

2 years ago

Which is the graph of 2x + 3y >-3?

Cerrena [4.2K]

Answer:

The third graph is correct for the inequality 2x + 3y > -3

Step-by-step explanation:

The slope of the graph is –2/3.

The y-intercept is –1

The area above the dashed be line is shaded.

To verify, choose a coordinate in the shaded area, substitute the values for x and y. Calculate and check.

For example: (1,1)

2(1) + 3(1) > –3

2 + 3 > –3

5 > –3 True.

4 0

2 years ago

4×-3(×-2)=21<br>Show your Work

IRINA_888 [86]

Answer:

24 does not equal 21

Step-by-step explanation:

4×-3(×-2)=21

Distribute:

4*6=21

24 does not equal 21

5 0

2 years ago

Ayuda necesito resolver este problema con procedimiento ;)

Paraphin [41]

$x^3-2x^2+x-1$ is one of the prime factors of the polynomial

<h3>How to factor the expression?</h3>

The question implies that we determine one of the prime factors of the polynomial.

The polynomial is given as:

$x^8 - 3x^6 + x^4 - 2x^3 - 1$

Expand the polynomial by adding 0's in the form of +a - a

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^7 - 4x^6 +x^6 + 2x^5 -2x^5- 3x^4 + 4x^4 + 2x^3 -6x^3+2x^3- x^2 -3x^2 +4x^2-2x+2x-1$

Rearrange the terms

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^5 - 3x^4 + 2x^3 - x^2 + 2x^7 - 4x^6 + 4x^4 -6x^3+4x^2-2x+x^6-2x^5+2x^3-3x^2+2x-1$

Factorize the expression

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^2(x^6-2x^5+2x^3-3x^2+2x-1) + 2x(x^6-2x^5+2x^3-3x^2+2x-1) + 1(x^6-2x^5+2x^3-3x^2+2x-1)$

Factor out x^6-2x^5+2x^3-3x^2+2x-1

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x^2+2x + 1)(x^6-2x^5+2x^3-3x^2+2x-1)$

Express x^2 + 2x + 1 as a perfect square

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6-2x^5+2x^3-3x^2+2x-1)$

Expand the polynomial by adding 0's in the form of +a - a

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^4-x^3 +x^3-2x^3-x^2 -2x^2 +x+x - 1)$

Rearrange the terms

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^3-x^4-2x^3-x^2+x+x^3-2x^2 +x - 1)$

Factorize the expression

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3(x^3-2x^2+x-1) -x(x^3-2x^2+x-1)+1(x^3-2x^2+x-1))$

Factor out x^3-2x^2+x-1

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3 -x+1)(x^3-2x^2+x-1)$

One of the factors of the above polynomial is $x^3-2x^2+x-1$ .

This is the same as the option (c)

Hence, $x^3-2x^2+x-1$ is one of the prime factors of the polynomial

Read more about polynomials at:

brainly.com/question/4142886

#SPJ1

4 0

1 year ago