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2 years ago

3 Write 3 as an improper fraction. 4

Mathematics

2 answers:

lyudmila [28]2 years ago

5 0

$\huge\boxed{Hi\: there!}$

3 as an improper fraction: 3/1.

Remember, an improper fraction is a fraction where the numerator is larger than the denominator.

$\huge\boxed{Therefore, \: 3 \: as \: an \: improper\: fraction\: is: \frac{3}{1} }$ $\huge\bold{Good \: luck\: with \: your\: studies! :)}\\\huge\bold{Answered \:by: \: An\: Unknown \: Dreamer}$

3241004551 [841]2 years ago

3 0

Answer:

I'm sorry, but this isn't clear. Maybe remake the layout you wrote? Because 3 3 4 doesn't make sense.

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270 american dollar and recevied 3000 pesos when he returned from mexico he had 100 pesos left. how many will he recevie when he

AURORKA [14]

To solve this problem we can use simple proportion
If
270$ -------------------------3000 pesos
x $ ---------------------------100 pesos (x$ means that we dont know how much)
Now we crossmultiplying to get proportion
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7 0

2 years ago

What is the volume in cubic centimeters of a right circular cone with radius 2 centimeters and height 9 centimeters?

andreev551 [17]

Answer:

$V = 37.7cm^3$

Step-by-step explanation:

The formula for the volume of a cone is:

$V = \frac{1}{3} \pi r^2h$

Given:

- Radius = 2cm

- Height = 9cm

Let's plug these into the equation:

$V = \frac{1}{3} \pi (2)^2(9)$

Now solve:

$V = \frac{1}{3} \pi (4)(9)\\V = \frac{1}{3}\pi 36\\V = \frac{36}{3} \pi \\V = 12\pi \\V = 37.7cm^3$

Therefore, the volume of the cone is 37.7cm^3.

<em>I hope this helps!!</em>

5 0

2 years ago

A researcher interested in the effects of humor on memory randomly assigns 16 participants to either the humor group or the no-h

masha68 [24]

Answer: option c

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3 0

3 years ago

A rectangular room is 22 meters longer than it is wide, and its perimeter is 1616 meters. find the dimension of the room.

Eddi Din [679]

W= width; L= length =w+22m; p= perimeter= 2(L+w)
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808m= 2w+22m subtract 22m from each side
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3 0

3 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