No hay solución en la que ambas trayectorias estén en la misma posición, puesto que no existe el riesgo de que sufran un choque entre ellos.
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La trayectoria del primero es:
Para el segundo, tiene-se que:
Igualando los valores de x:
No hay solución en la que ambas trayectorias estén en la misma posición, puesto que no existe el riesgo de que sufran un choque entre ellos.
Un problema similar es dado en brainly.com/question/24653364
5x = 2x + 9, where 5x > 0 and 2x + 9 > 0 ;
Then, 3x = 9 ;
Then, x = 3 ;
Verify : 5 × 3 = 2 × 3 + 9 ;correct !
5 × 3 > 0 ; correct !
2 × 3 + 9 > 0; correct !
Answer:
a)
We know that:
a, b > 0
a < b
With this, we want to prove that a^2 < b^2
Well, we start with:
a < b
If we multiply both sides by a, we get:
a*a < b*a
a^2 < b*a
now let's go back to the initial inequality.
a < b
if we now multiply both sides by b, we get:
a*b < b*b
a*b < b^2
Then we have the two inequalities:
a^2 < b*a
a*b < b^2
a*b = b*a
Then we can rewrite this as:
a^2 < b*a < b^2
This means that:
a^2 < b^2
b) Now we know that a.b > 0, and a^2 < b^2
With this, we want to prove that a < b
So let's start with:
a^2 < b^2
only with this, we can know that a*b will be between these two numbers.
Then:
a^2 < a*b < b^2
Now just divide all the sides by a or b.
if we divide all of them by a, we get:
a^2/a < a*b/a < b^2/a
a < b < b^2/a
In the first part, we have a < b, this is what we wanted to get.
Another way can be:
a^2 < b^2
divide both sides by a^2
1 < b^2/a^2
Let's apply the square root in both sides:
√1 < √( b^2/a^2)
1 < b/a
Now we multiply both sides by a:
a < b
The carrot patch is a 2 by 5 rectangle so 2 times 5 is 10 so the answer is A
the answer to your question is B.) 12x
