The absolute value of a negative number can be calculated by simply removing the negative symbol. Therefore,

\(\displaystyle \left | - 2 \frac{1}{2} \right | = 2 \frac{1}{2}\)

All four (negative) numbers in the set \(\displaystyle \left \{ -4, -3, -2, -1 \right \}\) are less than this positive number.

Let's reductively consider what this data tells us.

Consider each group (a,b,c) as a group of signs.

From abc < 0, we know that the following are possible:

(–, +, +), (+, –, +), (+, +, –), (–, –, –)

From ab > 0, we know that we must eliminate (–, +, +) and (+, –, +)

From bc > 0, we know that we must eliminate (+, +, –)

Therefore, any of our answers must hold for (–, –, –)

This eliminates immediately a > 0, b > 0

Likewise, it eliminates a – b > 0 because we do not know the relative sizes of a and b. This could therefore be positive or negative.

Finally, ac is a product of negatives and is therefore positive. Hence ac < 0 does not hold.

We are left with a + b < 0, which is true, for two negatives added must be negative.

A negative number divided by a negative number always results in a positive number. \(\displaystyle 36\) divided by \(\displaystyle 9\) equals \(\displaystyle 4\). Since the answer is positive, the answer cannot be \(\displaystyle -4\) or any other negative number.

Begin by isolating your variable.

Subtract \(\displaystyle x\) from both sides:

\(\displaystyle 16-4x-x=6\), or \(\displaystyle 16-5x=6\)

Next, subtract \(\displaystyle 16\) from both sides:

\(\displaystyle -5x=6-16\), or \(\displaystyle -5x=-10\)

Then, divide both sides by \(\displaystyle -5\):

\(\displaystyle x=\frac{-10}{-5}\)

Recall that division of a negative by a negative gives you a positive, therefore:

\(\displaystyle x=\frac{10}{5}\) or \(\displaystyle x=2\)

Because \(\displaystyle \dpi{100} \small -3b\) is positive, \(\displaystyle \dpi{100} \small b\) must be negative since the product of two negative numbers is positive.

Because \(\displaystyle \dpi{100} \small ab\) is also positive, \(\displaystyle \dpi{100} \small a\) must also be negative in order to produce a prositive product.

To check you answer, you can try plugging in any negative number for \(\displaystyle \dpi{100} \small a\).

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.