the length of a rectangle is 9 centimeters more than the width. the area is 112 square centimeters. find the length and width of the rectangle

Answers

Answer 1

Answer: l = w + 9

lw = 112

Substitute l = w + 9 into the second equation

(w + 9)(w) = 112

w^2 + 9w - 112 = 0
(w+16)(w-7) = 0

w = -16
or
w = 7

we have to take the positive value, so w = 7

l = w + 9
l = 7 + 9
l = 16

The length is 16, and the width is 7

Answer 2

Answer: here,
let width(b)be x then,
length (l)=9cm+x
area =112 sq cm
now,
area of rectangle=l*b
or, 112=(9+x)x
or, 112=9x+x^2
or, 0=x^2+9x-112
or, 0=x^2+(16-7)x-112
or, 0=x^2+16x-7x-112
or, 0=x(x+16)-7(x+16)
or, 0=(x-7)(x+16)
either,
0=x-7
or,7=x
x=7cm
Or,
0=x+16
or, -16=x
x= -16[impossible,as distance is never negative] so,
x=7cm
therefore,length = 7cm + 9 cm = 16cm and width = 7cm.
;)

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Solve for x 5x−1=6x−9 It would be better for more of a simpler and easier method if possible.

Answers

Answer:

X=8

Step-by-step explanation:

$5x - 1 = 6x - 9$

MovevariabletoL.H.Sandchangeitssign:

$5x - 6x - 1 = - 9$

MoveconstanttoRHSandchangeitssign:

$5x - 6x = - 9 + 1$

Combinelike termsandsimplify:

$- x = - 8 \n x = 8$

Hopethishelps....

Goodluck on your assignment...

Since the creation of a social program, the percent of persons 65 years and over with a family income below the poverty level has declined. The percent can be approximated by the function P(t)=30.71-5.77ln t, where t is the number of years since 1965. Find the percent of persons 65 & over with family income below the poverty level and the rate of change in 1970.

i know how to find the percent, but don't know how to find the rate of change. Please help!