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2 years ago

I NEED HELP PLEASE directions:Work and Solution

Mathematics

1 answer:

Volgvan2 years ago

4 0

Answer:

2(x + 4) / 6(x² - 3x - 28)

Step-by-step explanation:

Area of a rectangle = length × width

Length = 2/(x² - 3x - 28)

Width = x² - 16/6x - 24

= (x + 4)(x - 4) / 6(x - 4)

= (x + 4) / 6

Area of a rectangle = length × width

= 2/(x² - 3x - 28) × (x + 4) / 6

= 2(x + 4) / (x² - 3x - 28)6

= 2(x + 4) / 6x² - 18x - 168

= 2(x + 4) / 6(x² - 3x - 28)

Area of a rectangle =

2(x + 4) / 6(x² - 3x - 28)

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A factory packed 9600kg of flour in to small bags and large bags. Each small bag contained 3kg of flour and each large bag conta

yawa3891 [41]

Answer:

696 large bags of flour were packed.

Step-by-step explanation:

Let large bags of flour be 'y'

Let small bags of flour be 'x'. That is x=2040 (according to the question)

$(3 \times x) + (5 \times y) = 9600$

$(3 \times 2040) + ( 5 \times y) = 9600$

$(6120) + (5 \times x) = 9600$

$(5 \times x) = 9600 - 6120$

$(5 \times x) = 3480$

$x = 3480 \div 5$

$x = 696$

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3 years ago

Write an equation for the situation then solve for x. After paying $7 for a sandwich, John has $11. With how much money did he s

yKpoI14uk [10]

Answer:

x = 18

Step-by-step explanation:

total money = money spent + money left

x = 7+11

x = 18

8 0

3 years ago

Which function has the following characteristics?

iren [92.7K]

Using the asymptote concept, the function with a vertical asymptote at x = 3 and an horizontal asymptote at $y = -\frac{1}{2}$ is given by:

$f(x) = -\frac{x}{2(x - 3)}$

<h3>What are the asymptotes of a function f(x)?</h3>

The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.

The horizontal asymptote is the value of f(x) as x goes to infinity, as long as this value is different of infinity.

The vertical asymptote at x = 3 means that x = 3 is a root of the denominator, hence:

$f(x) = \frac{g(x)}{x - 3}$

The horizontal asymptote at y = -1/2 means that:

$\lim_{x \rightarrow \infty} f(x) = -\frac{1}{2}$

Which happens if $g(x) = -\frac{x}{2}$ , hence the function is:

$f(x) = -\frac{x}{2(x - 3)}$

More can be learned about asymptotes at brainly.com/question/16948935

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2 years ago

To make a shade of lavender paint jon mixes 4 ounces of red tint and 28 ounces of blue tint into one gallon of white paint. if 2

kirill [66]

You can write this as a ratio to find the answer, but you first need to convert the gallon of white paint into ounces, which is 160 ounces. The ratio, red to blue to white, is 4:28:160, which you can simplify by dividing by 4 into 1:7:40. If 80 ounces of red tint are used, you need to multiply each value in the ratio by 80, which gives you 80:560:3200. This shows that you need 560 ounces of blue tint, which can also be written as 3.5 gallons. I hope this helps!

8 0

3 years ago

Can anyone figure this out?

Verizon [17]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill$

$\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}$

now that we know how long each one is, let's plug those in Heron's Area formula.

$\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18$

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3 years ago