A negative integer raised to an integer power is positive if and only if the absolute value of the exponent is even. Since the sum or difference of two odd integers is always an even integer, this is the case in all three expressions. The correct response is all of these.

A negative integer raised to an integer power is positive if and only if the absolute value of the exponent is even; it is negative if and only if the absolute value iof the exponent is odd. Therefore, all three expressions have signs that are dependent on the odd/even parity of \(\displaystyle a\) and \(\displaystyle c\), which are not given in the problem. 

The correct response is none of these.

The key is knowing that a negative number raised to an odd power yields a negative result, and that a negative number raised to an even power yields a positive result.

\(\displaystyle \frac{ a^{6}b^{4}}{c^{9}}\): \(\displaystyle a^{6}\) and \(\displaystyle b^{4}\) are positive, yielding a positive dividend; \(\displaystyle c^{9}\) is a negative divisor; this result is negative.

\(\displaystyle \frac{ a^{7}c^{5}}{b^{3}}\): \(\displaystyle a^{7}\) and \(\displaystyle c^{5}\) are negative, yielding a positive dividend; \(\displaystyle b^{3}\) is a negative divisor; this result is negative.

\(\displaystyle \frac{ a^{8}c^{3}}{b^{6}}\) : \(\displaystyle a^{8}\) is positive and \(\displaystyle c^{3}\) is negative, yielding a negative dividend; \(\displaystyle b^{3}\) is a positive divisor; this result is negative.

\(\displaystyle \frac{ b^{5}c^{4}}{a^{12}}\) : \(\displaystyle b^{5}\) is negative and \(\displaystyle c^{4}\) is positive, yielding a negative dividend; \(\displaystyle a^{12}\) is a positive divisor; this result is negative.

\(\displaystyle \frac{a^{4}b^{7}}{c^{5}}\): \(\displaystyle a^{4}\)  is positive and \(\displaystyle b^{7}\) is negative, yielding a negative dividend; \(\displaystyle c^{5}\) is a negative divisor; this result is positive.

The correct choice is \(\displaystyle \frac{a^{4}b^{7}}{c^{5}}\).

Since \(\displaystyle a\) and \(\displaystyle c\) are positive, all powers of \(\displaystyle a\) and \(\displaystyle c\) will be positive; also, in each of the expressions, the powers of \(\displaystyle a\) and \(\displaystyle c\) are being added. The clue to look for is the power of \(\displaystyle b\) and the sign before it.

In the cases of \(\displaystyle a^{3}+ b^{4} +c\) and \(\displaystyle a+ b^{2} +c^{6}\), since the negative number \(\displaystyle b\) is being raised to an even power, each expression amounts to the sum of three positive numbers, which is positive.

In the cases of \(\displaystyle a^{2} -b^{7} +c\) and \(\displaystyle a^{2}-b^{3} +c\), since the negative number \(\displaystyle b\) is being raised to an odd power, the middle power is negative - but since it is being subtracted, it is the same as if a positive number is being added. Therefore, each is essentially the sum of three positive numbers, which, again, is positive.

In the case of \(\displaystyle a^{4}+ b^{5} +c^{3}\), however, since the negative number \(\displaystyle b\) is being raised to an odd power, the middle power is again negative. This time, it is basically the same as subtracting a positive number. As can be seen in this example, it is possible to have this be equal to a negative number:

\(\displaystyle a = 1, b = -2, c= 1\):

\(\displaystyle a^{4}+ b^{5} +c^{3} = 1^{4}+ (-2)^{5} +1^{3} = 1 + (-32)+ 1 = -30\)

Therefore, \(\displaystyle a^{4}+ b^{5} +c^{3}\) is the correct choice.

Since \(\displaystyle b^{2}\) is positive, \(\displaystyle ab^{2}\), the product of a negative number and a positive number, must be negative also.

Of the others:

\(\displaystyle a^{2}b\) is incorrect; if \(\displaystyle a\) is negative, then \(\displaystyle a^{2}\) is positive, and \(\displaystyle a^{2}b\) assumes the sign of \(\displaystyle b\).

\(\displaystyle a^{2}+ b\) is incorrect; again, \(\displaystyle a^{2}\) is positive, and if \(\displaystyle b\) is a positive number, \(\displaystyle a^{2}+ b\) is positive.

\(\displaystyle a+b^{2}\) is incorrect; regardless of the sign of \(\displaystyle b\), \(\displaystyle b^{2}\) is positive, and if its absolute value is greater than that of \(\displaystyle a\), \(\displaystyle a+b^{2}\) is positive.

If \(\displaystyle -10 \leq a \leq -1\), then we know that \(\displaystyle a\) is any number between or equal to \(\displaystyle -10\) and \(\displaystyle -1\). Therefore \(\displaystyle a\) must be a negative number. 

Also, if \(\displaystyle -10 \leq b \leq -1\), then we know that \(\displaystyle b\) is any number between or equal to \(\displaystyle -10\) and \(\displaystyle -1\). Therefore \(\displaystyle b\) must be a negative number.

Now looking the expression \(\displaystyle (a -10)(b +10)\) we can find the sign of each component in the expression.

Since \(\displaystyle a\) is negative, we know that a negative number minus another number is still a negative number.

Therefore, \(\displaystyle (a-10)\) is a negative number.

Since \(\displaystyle b\) is between or equal to  \(\displaystyle -10\) and \(\displaystyle -1\) we can plug in these end values in to determine the sign of \(\displaystyle (b +10)\).

\(\displaystyle (-10+10)=0\)

\(\displaystyle (-1+10)=9\)

Therefore, \(\displaystyle (b +10)\) is either zero or a positive number.

Now to find the sign of the expression we look at the product of the two components. The product of a negative number and a positive number is a negative number; the product of a negative number and zero is zero. Therefore, the correct choice is that \(\displaystyle (a -10)(b +10)\) is negative or zero.

When multiplying together two negatives, our value for the product become positive. \(\displaystyle (-270) \times (-131) = 270 \times 131 = 35370\)

Since we have one positive and one negative multiple, the resulting product must be negative. \(\displaystyle 76 \times (-31) = -(76\times 31) = -2356\)

By choosing a random negative number, for example: –4, we can input the number into each choice and see if we come out with another negative number.  When we put –4 in for x, we would have –4 – (–(–4)) or –4 – 4, which is –8.  Plugging in the other options gives a positive answer.  You can try other negative numbers, if needed, to confirm this still works. 

x: –7 – 7= –7 + –7 = –14

y: –7 – (–7) = –7 + 7 = 0

when subtracting a negative number, turn it into an addition problem