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2 years ago

- 4x +35> 2 What is the answer?

Mathematics

1 answer:

11111nata11111 [884]2 years ago

8 0

Answer:

x < $\frac{33}{4}$

Step-by-step explanation:

Given

- 4x + 35 > 2 ( subtract 35 from both sides )

- 4x > - 33

Divide both sides by - 4, reversing the symbol as a result of dividing by a negative quantity.

x < $\frac{33}{4}$

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HELP!! I desperately need help!! I would appreciate if you could show work please!!!

zzz [600]

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a) 137.9°

b) 12.2° (please refer to explanation because I didn't know what to round to)

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1a) A supplementary angle will equal 180°, so you must subtract 42.1 from 180

180-42.1 = 137.9

1b) A complementary angle will equal 90°, so you need to use the expression given and add an x for the other part of 90°

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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