Answer: 10, 11, & 12
<u>Step-by-step explanation:</u>
Let x represent the age of the youngest child.
Their ages are consecutive so,
Youngest: x
Middle: x + 1
Oldest: x + 2
The age of the Youngest squared (x²) equals 8 times the Oldest [8(x + 2)] plus 4.
x² = 8(x + 2) + 4
x² = 8x + 16 + 4
x² = 8x + 20
x² - 8x - 20 = 0
(x - 10)(x + 2) = 0
x - 10 = 0 or x + 2 = 0
x = 10 or x = -2
Since age cannot be negative, x = -2 is not valid
So, the Youngest (x) is 10
the Middle (x + 1) is 11
and the Oldest (x + 2) is 12
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11. You’ve done it correctly
12. Let x^2=y
y^2+13y+40=0
(y+8)(y+5)=0
y=8, 5
Since y=x^2
x^2=8 x^2=5
x=+/-√5 x= +/-2√2
13. x^4-x^2-x^2-8=0
x^4-2x^2-8=0
let x^2=y
Y^2-2y-8=0
(y-4)(y+2)=0
y=4, -2
Since y=x^2
X^2=4 X^2=-2
X= +/- 2 This wouldn’t be a real solution
14. It’s pretty much the same process, just substitute y in for x^2. If you’re confused feel free to ask and I can do it, or you can put it through Photomath
15. You’re on the right track so I’m just going to continue from where you left off
x^2(4x+5)-4(x+5)=0
(x^2-4)(4x+5)=0
x= +/- 2 4x=5
x=5/4 or 1 1/4
Hope this helped :)
Answer:
16.5 ft by 25.5 ft
Step-by-step explanation:
Let w represent the width of the garden in feet. Then w+9 is the garden's length, and w(w+9) represents its area.
The surrounding walkway adds 8 feet to each dimension, so the total area of the garden with the walkway is ...
(w+8)(w+9+8) = w^2 +25w +136
If we subtract the area of the garden itself, then the remaining area is that of the walkway:
(w^2 +25w +136) - (w(w+9)) = 400
16w + 136 = 400 . . .simplify
16w = 264 . . . . . . . . subtract 136
264/16 = w = 16.5 . . . . . width of the garden in feet
w+9 = 25.5 . . . . . . . . . . .length of the garden in feet
