How many $10$-digit numbers are there, such that the sum of the digits is divisible by $2$?

Complete question is;

How many 10 - digit numbers are there, such that the sum of the digits is divisible by 2?

There are 4500000000 ten digit numbers whose sum is divisible by 2.

Step-by-step explanation:

Since we are dealing with 10 digit numbers, the first 10 digit number whose sum is divisible by 2 is; 1000000001.

While the last 10 digit number whose sum is divisible by 2 is 9999999999.

Now,to find how many 10 digit numbers whose sum are divisible by 2, we will simply divide the difference of both numbers by 2 and then add 1 to the answer. This is because subtracting the minimum from the maximum gives us the number of 10 digit numbers we have. Then dividing by 2 gives the number of ten digit numbers with sum divisible by 2 in between. Then adding 1 to the result to cover the number not included gives;

((9999999999 - 1000000001)/2) + 1 = 4500000000

d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204