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3 years ago

Someone be as kind as to offer some help?

Mathematics

1 answer:

RideAnS [48]3 years ago

4 0

Answer:

x = 4 and y = 0

Step-by-step explanation:

Using the elimination method:

4(-x + 5y=-4) --> -4x + 20y = -16.

Now we add -4x + 20y = 016 and 4x + 3y = 16. This gives us 23y = 0 or just y =0.

Now we can find x by plugging in y.

-4x + 0 = -16.

Divide by -4 on both sides -> x = 4.

If you want to double check you can plug your values into -4x + 20y = -16.

-4(4) +20(0) = -16.

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<em>In order to increase the area of a rectangular garden that measures 12 feet by 14 feet by 50% Jake must increase each dimension by equal lengths, x:</em>

$x\approx 2.9ft$

<h2>Explanation:</h2><h2 />

First of all, let's calculate the area of the original rectangular garden:

$A=b\times h \\ \\ b:base \\ \\ h:height \\ \\ \\ b=12ft \\ \\ h=14ft \\ \\ \\ A=12(14) \\ \\ A=168ft^2$

Jake wants to increase the area by 50%, so the new area would be:

$A'=168(1.5) \\ \\ A'=252ft^2$

He wants to increase the area by 50% and plans to increase each dimension by equal lengths, x, so this is represented by the figure below, therefore:

$(12+x)(14+x)=252 \\ \\ 168+12x+14x+x^2-252=0 \\ \\ x^2+26x-84=0 \\ \\ \\ Using \ quadratic \ formula: \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ a=1 \\ \\ b=26 \\ \\ c=-84 \\ \\ \\ x=\frac{-26 \pm \sqrt{26^2-4(1)(-84)}}{2(1)} \\ \\ x=\frac{-26 \pm \sqrt{1012}}{2} \\ \\ \\ Two \ solutions: \\ \\ x_{1}=-13+\sqrt{253} \approx 2.9\\ \\ x_{2}=-13-\sqrt{253} \approx -28.9 \\ \\ x_{2} \ is \ discarded \ because \ it \ can't \ be \ negatives$

Finally:

<em>In order to increase the area of a rectangular garden that measures 12 feet by 14 feet by 50% Jake must increase each dimension by equal lengths, x:</em>

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<h2>Learn more:</h2>

Dilation: brainly.com/question/10945890

#LearnWithBrainly

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