Let the width path be x.
Length of the outer rectangle = 26 + 2x.
Width of the outer rectangle = 8 +2x.
Combined Area = (2x + 26)*(2x + 8) = 1008
2x*(2x + 8) + 26*(2x + 8 ) = 1008
4x² + 16x + 52x + 208 = 1008
4x² + 68x + 208 - 1008 = 0
4x² + 68x - 800 = 0. Divide through by 4.
x² + 17x - 200 = 0 . This is a quadratic equation.
Multiply first and last coefficients: 1*-200 = -200
We look for two numbers that multiply to give -200, and add to give +17
Those two numbers are 25 and -8.
Check: 25*-8 = -200 25 + -8 = 17
We replace the middle term of +17x in the quadratic expression with 25x -8x
x² +17x - 200 = 0
x² + 25x - 8x - 200 = 0
x(x + 25) - 8(x + 25) = 0
(x+25)(x -8) = 0
x + 25 = 0 or x - 8 = 0
x = 0 -25 x = 0 + 8
x = -25 x = 8
The width of the path can not be negative.
The only valid solution is x = 8.
The width of the path is 8 meters.
Answer:
A) 21/7, or 3
B) 36/5, or 7 1/5
C) 35/25, or 1 15/25, or 1 3/5
D) 54/12, or 4 6/12, or 4 1/2
Step-by-step explanation:
8/7 plus 13/7. You add the numerators (the numbers on top) together. That makes 21. The denominator (the numbers on bottom) stays the same. So it would be 21/7. 7 fits into 21, 3 times.
8/7 más 13/7. Agrega los numeradores (los números en la parte superior) juntos. Eso hace 21. El denominador (los números en la parte inferior) permanece igual. Entonces sería 21/7. 7 encaja en 21, 3 veces.
8/7 plus 13/7. Vous ajoutez les numérateurs (les chiffres du haut) ensemble. Cela fait 21. Le dénominateur (les chiffres du bas) reste le même. Donc, ce serait 21/7. 7 correspond à 21, 3 fois.
8/7 plus 13/7. Quarum numeratores addere (supra de numero) una. 21. Quod facit denominator est (per numeros in fundo) manebit. Ita esset 21/7. Vicium, in VII XXI, III tempora.
Answer:
D I think
Step-by-step explanation:
Im not sure what the question is asking but my guess would be c)18.
Answer:
On the surface, it seems easy. Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y = -1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42?
That turned out to be much harder—as in, no one was able to solve for those integers for 65 years until a supercomputer finally came up with the solution to 42. (For the record: x = -80538738812075974, y = 80435758145817515, and z = 12602123297335631. Obviously.)
Step-by-step explanation:
