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2 years ago

4x ≤12 dimplify solve for x

Mathematics

2 answers:

juin [17]2 years ago

8 0

Answer:

$\boxed{x\leq 3}$

Step-by-step explanation:

$4x \leq 12$

$\sf Divide \ both \ parts \ by \ 4.$

$\displaystyle \frac{4x}{4} \leq \frac{12}{4}$

$x\leq 3$

EastWind [94]2 years ago

7 0

Answer:

x<3

Step-by-step explanation:

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Lisa ran y miles on monday and 3 times as far on tuesday . write an expression in simplest form to sho the number of miles she r

never [62]

Answer:

y=3x

Step-by-step explanation:

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2 years ago

Charlie watched 1 3 of a movie in the morning, and he watched some more at night. By the end of the day, he still had 2 5 of the

Finger [1]

Answer: 4/15 of the movie

Step-by-step explanation:

In the morning, Charlie watched ¹/₃ of a movie. He then watched some more at night.

At the end of the day, he still had ²/₅ of the movie to watch.

The proportions of an item must always add up to one. This means that if you added the 1/3 to the 2/5 and then to the fraction Charlie watch at night, they should add up to one.

¹/₃ + ²/₅ + x = 1

x = 1 - ¹/₃ - ²/₅

$x = \frac{15 - 5 - 6 }{15} \\\\x = \frac{4}{15}$

3 0

2 years ago

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Lena [83]

The answer is the second sentence

4 0

1 year ago

What is the square root of A=121 ft2

storchak [24]

The result of the square root of the equation A = 121 ft² is √A = 11 ft

<h3>What are square roots?</h3>

Square roots are numbers multiplied by itself to give another number

<h3>How to determine the square root?</h3>

The equation is given as:

A = 121 ft²

Take the square root of both sides

√A = √121 ft²

Evaluate the square root

√A = 11 ft

Hence, the square root of the equation A = 121 ft² is √A = 11 ft

Read more about square roots at:

brainly.com/question/98314

#SPJ1

4 0

1 year ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago