A few days ago

# math homework please help…?

> if a and b are positive consecutive integers, where B > A, which of the following is equal to A^2 + B^2?

(A) A^2 + 2

(B) A^2 + 2A

(C) A^2 + A + 2

(D) 2(a^2 + 2a + 2)

(E) 4(a^2 + a + 1)

2> For how many ordered pairs of positive integers (x, y) is

3X + 5Y less than or equal to 16?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

thanks in advance!

Top 1 Answers

A few days ago

Favorite Answer

For the first one, you need to rewrite b in terms of a. Since they’re consecutive integers, b is one more than a. That is, b = a + 1.

Now, plug the equivalent of b into the expression a^2 + b^2:

a^2 + (a + 1)^2

= a^2 + (a^2 + 2a + 1)

= 2a^2 + 2a + 1

So none of those answers are correct.

For the second one, find the highest value of x for which this is true:

Again, rewrite y as x + 1.

3x + 5(x + 1) < 16 8x + 5 < 16 8x < 11 Looks like this is only true for x = 1. So, you get one ordered pair: x = 1 and y = 2.

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