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2 years ago

Which value is a solution to 3y + 4 < 16?

Mathematics

1 answer:

bixtya [17]2 years ago

8 0

Hello,

3y + 4 < 16

⇔ 3y < 16 - 4

⇔ 3y < 12

⇔ y < 12/3

⇔ y < 4

⇒ S = ] - ∞ ; 4 [

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andrew-mc [135]

The remainder from the division of the algebraic equation is -53/8.

<h3>What is the remainder of the algebraic expression?</h3>

The remainder of the algebraic expression can be determined by using the long division method.

Given that:
$\mathbf{f(x) = \dfrac{x^3 - 6x^2 + 3x - 1}{2x-3}}$

where:

The divisor = 2x -3

Using the long division method, we have:

$\mathbf{= \dfrac{x^2}{2} +\dfrac{-\dfrac{9x^2}{2}+3x -1 }{2x-3}}$

$\mathbf{= \dfrac{x^2}{2}-\dfrac{9x}{4} +\dfrac{-\dfrac{-15x}{4}-1 }{2x-3}}$

$\mathbf{= \dfrac{x^2}{2}-\dfrac{9x}{4} -\dfrac{15}{8}+\dfrac{-\dfrac{53}{8} }{2x-3}}$

$\mathbf{= \dfrac{x^2}{2}-\dfrac{9x}{4} -\dfrac{15}{8}-\dfrac{53 }{8(2x-3)}}$

Therefore, we can conclude that the remainder is -53/8.

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brainly.com/question/4541471

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Which expression is equivalent to StartFraction (3 m Superscript negative 2 Baseline n) Superscript negative 3 Baseline Over 6 m

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Option a: $\frac{m^{5} }{162n}$ is the equivalent expression.

Explanation:

The expression is $\frac{(3m^{-2} n)^{-3}}{6mn^{-2} }$ where $m\neq 0, n\neq 0$

Let us simplify the expression, to determine which expression is equivalent from the four options.

Multiplying the powers, we get,

$\frac{3^{-3}m^{6} n^{-3}}{6mn^{-2} }$

Cancelling the like terms, we have,

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This equation can also be written as,

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Multiplying the terms in denominator, we have,

$\frac{m^{5} }{162n}$

Thus, the expression which is equivalent to $\frac{(3m^{-2} n)^{-3}}{6mn^{-2} }$ is $\frac{m^{5} }{162n}$

Hence, Option a is the correct answer.

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