A rectangle is 1 inch longer than it is wide. Its diagonal is 5 inches. What's the width of the rectangle?

Step-by-step explanation:

Let x be the length and y be the width

The length is 1 inch longer than the width

So : x-1 =y

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The diagonal is 5 inches

The diagonal, width and length are forming a right triangle where the diagonal is the hypotenus

The Pythagorian theorem:

x^2 + y^2 = 5^2

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Let's calculate y the width

The system is :

x-1 = y => x =y+1

x^2 + y^2 = 25

Replace x in the second equation with y+1 :

(y+1)^2 + y^2 = 25

y^2 + 2y + 1 + y^2 = 25

2y^2 + 2y +1 = 25

2y^2 + 2y -24 = 0

This is a quadratic equation so we will use the discriminant

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the discriminat is b^2-4ac

● b =2

● a = 2

● c = -24

b^2 -4ac = 4-4*2(-24) =196 > 0

Since the dicriminant is positive this equation has two solutions y and y'

y = (-2-14)/4 = -16/4 = -4

y '= (-2+14)/4 = 12/4 = 3

The first solution is negative

A distance is always positive so the first solution can't be the width

Then the width is 3

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

f(x) is going down by 4

step-by-step explanation:

(its either that or its going left by 4)