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3 years ago

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A track practice, sheila worked on the long jump.her second jump was measured 2 yards, 1 foot,10 inches.what was the total lengt

h of sheila's two jump express your answer in inches.

1 answer:

zaharov [31]3 years ago

3 0

94 inches

hope this helps

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The length of a rectangle is 5 inches more than its vedth. The area of the rectangle is equal to

jarptica [38.1K]

Answer:

The width = x = 14

The length = x+5 = 14+5 = 19

Step-by-step explanation:

Let l = length, w = width, A = area = l×w , and P = perimeter = 2(l+w)

Let 'x' be the width

As the length 'l' of a rectangle is 5 inches more than its width.

so the length will be = x+5

As the Area of the rectangle is equal to 2 inches more than 4 times the perimeter.

A = 4P + 2

so the equation becomes

l × w = 4×2(l+w)+2

substituting w=x, l = x+5,

(x+5)x = 8(x+5+x)+2

x²+5x = 8(5+2x)+2

x²+5x = 40+16x+2

x²+5x = 16x+42

x²+5x-16x-42 =0

x²-11x-42=0

$x^2-11x=42$

$\mathrm{Add\:}a^2=\left(-\frac{11}{2}\right)^2\mathrm{\:to\:both\:sides}$

$x^2-11x+\left(-\frac{11}{2}\right)^2=42+\left(-\frac{11}{2}\right)^2$

$x^2-11x+\left(-\frac{11}{2}\right)^2=\frac{289}{4}$

$\left(x-\frac{11}{2}\right)^2=\frac{289}{4}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{11}{2}=\sqrt{\frac{289}{4}}$

$x-\frac{11}{2}=\frac{\sqrt{289}}{\sqrt{4}}$

$x-\frac{11}{2}=\frac{\sqrt{289}}{2}$

$x-\frac{11}{2}=\frac{17}{2}$

$x-\frac{11}{2}+\frac{11}{2}=\frac{17}{2}+\frac{11}{2}$

$x=14$

similarly solving

$x-\frac{11}{2}=-\sqrt{\frac{289}{4}}$

$x-\frac{11}{2}=-\frac{17}{2}$

$x-\frac{11}{2}+\frac{11}{2}=-\frac{17}{2}+\frac{11}{2}$

$x=-3$

so

x = 14, or x = -3

As the width 'x' can not be negative.

so x = 14

Thus,

The width = x = 14

The length = x+5 = 14+5 = 19

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