Answer:
What if we made a fast game?
Explanation:
The mascot is the person, animal or thing which brings good luck, just like Mario. However, this is not a problem with Sonic the Hedgehog. And the Hedgehog is powerful enough to cross any barrier. Hence tieing the collectibles is not required. And the Hedgehog is quite cute and domestic. But it moves only 10 meters per second if we are considering the maximum speed. Hence, the correct option is certainly what if we made a fast game? And this is option C.
Respuesta: Los caracteres adquiridos no se transmiten genéticamente porque no modifican el ADN de los organismos
Explicación:
Jean-Baptiste Lamarck al igual que Charles Darwin, propuso una teoría sobre la evolución que explicaba cambios en los organismos a través del tiempo. La teoría de Lamarck se enfocaba en condiciones en el ambiente que propiciaban cambios en los organismos. Un ejemplo de esto son las jirafas, que de acurdo a Lamarck tenían cuellos largos debido al esfuerzo continuado para comer hojas de árboles altos. Esto significa que la característica de cuello largo era adquirido por las jirafas durante su vida y según Lamarck se transmitiría a sus descendientes.
Sin embargo, se ha comprobado que los caracteres adquiridos no modifican el ADN de los organismos, por ejemplo las cirugías estéticas no cambian el ADN de una persona y por esta razón no son transmitidos a sus descendientes. Por el contrario, en las poblaciones de organismos ciertas características prevalencen en el tiempo debido a la selección natural. Esto significa que el cuello de las jirafas es el resultado que el cuello largo sea una característica beneficiosa que ha prevalecido debido a la selección natural y no de características adquiridas que son transmitidas a descendientes.
I hope this helps but cars, phones and computers.
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
