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3 years ago

The perimeter of a soccer field is 300 yards. if the length is 50 yards longer than the width, what are the field's dimensions?

Mathematics

2 answers:

Ket [755]3 years ago

6 0

I'm pretty sure the answer is L = 100
(:

Marrrta [24]3 years ago

5 0

X(x+50)=300

x2 +50x - 300=0

use quadratic formula to get x

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algol [13]

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Solve for x.

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3 years ago

Which of the following are solutions to the equation below?

sladkih [1.3K]

Answer:

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

Step-by-step explanation:

considering the equation

$2x^2\:-\:4x\:-\:3\:=\:x$

solving

$2x^2\:-\:4x\:-\:3\:=\:x$

$2x^2-4x-3-x=x-x$

$2x^2-5x-3=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=-5,\:c=-3:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

solving

$x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

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$x=\frac{5+7}{4}$

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also solving

$x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5-\sqrt{49}}{4}$

$x=-\frac{2}{4}$

$x=-\frac{1}{2}$

Therefore,

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

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3 years ago

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irga5000 [103]

Answer: 5.36363636364

Step-by-step explanation:

91+ 129+ 16=236

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