Algebra - PSAT Math

The equation simplifies to x > -1.41. -1 is the answer.

Let's take this step by step. "Half of one hundred" is 100/2 = 50. Then "half of one hundred divided by five" is 50/5 = 10. "Multiplied by one-tenth" really is the same as dividing by ten, so the last step gives us 10/10 = 1.

We are told that x&y = (x – y)xy .

–1&2 = (–1 – 2)(–1)(2) = (–3)–2

To simplify this, we can make use of the property of exponents which states that a– b = 1/(ab ).

(–3)–2 = 1/(–3)2 = 1/9

The answer is 1/9.

The one-time first-month cost is p, and the monthly cost is m, which gets multipled by every month but the first (of which there are n -1). The total cost is the first-month cost of p, plus the monthly cost for (i.e. times) n -1 months, which makes the total cost equal to p + m (n -1).

Plug each value for x into the above equation and solve for f(x). f(1) provides the lowest value –5/3

Recall that with absolute values and "less than" inequalities, we have to hold the following:

12x + 3y < 15

AND

12x + 3y > –15

Otherwise written, this is:

–15 < 12x + 3y < 15

In this form, we can solve for y. First, we have to subtract x from all 3 parts of the inequality:

–15 – 12x < 3y < 15 – 12x

Now, we have to divide each element by 3:

(–15 – 12x)/3 < y < (15 – 12x)/3

This simplifies to:

–5 – 4x < y < 5 – 4x

This inequality could be rewritten as:

4x + 14 > 30 OR 4x + 14 < –30

Solve each for x:

4x + 14 > 30; 4x > 16; x > 4

4x + 14 < –30; 4x < –44; x < –11

Therefore, anything between –11 and 4 (inclusive) will not work. Hence, the answer is 7.

To solve this problem, we look for an operation to perform on both sides that will leave n-2 by itself on one side. Dividing both sides by 25_n_-3 would leave n-2 by itself on the right side of the equqation, as shown below:

100n3p–1/25n–3 = 25n/25n–3

Remember that when dividing terms with the same base, we subtract the exponents, so the equation can be written as 100n0p–1/25 = n–2

Finally, we simplify to find 4_p–_1 = _n–_2.

Adding 12 to both sides and dividing by 4 yields (7y+12)/4.

By distribution we obtain **–**7x – 21 = – 7x + 20. This equation is never possibly true.

For this problem, we must take into account the absolute value.

First, we solve for 2_x_ – 2 > 20. But we must also solve for 2_x_ – 2 < –20 (please notice that we negate 20 and we also flip the inequality sign).

First step:

2_x_ – 2 > 20

2_x_ > 22

x > 11

Second step:

2_x_ – 2 < –20

2_x_ < –18

x < –9

Therefore, x > 11 and x < –9.

A possible value for x would be –10 since that is less than –9.

Note: the value 11 would not be a possible value for x because the inequality sign given does not include an equal sign.

Subtract from both sides:

Subtract 7 from both sides:

Divide both sides by :

Remember to switch the inequality when dividing by a negative number:

Since is not an answer, we must find an answer that, at the very least, does not contradict the fact that is less than (approximately) 4.67. Since any number that is less than 4.67 is also less than any number that is bigger than 4.67, we can be sure that is less than 5.

We need to rewrite this problem in mathematic terms.

8 less than 32 can be written as 32 – 8

so we will get the equation

18 – w = 32 – 8.

Now, we need to solve for w.

–w = 32 – 8 – 18

–w = 6

w = –6

Find the value of the given expression, , by plugging in –6 for w.

For F(1/2), x2=1/4, which is less than 1, so we use the bottom equation to solve. This gives F(1/2)= 4(1/2)2 – 2(1/2) + 2 = 1 – 1 + 2 = 2

Subtract 5 from both sides of the inequality:

Multiply both sides by 5:

Therefore only I must be true.

Add the two equations together to yield 5x + 5y = 9a + b, then factor out 5 to get 5(x + y) = 9a + b; divide both sides by 5 to get x + y = (9a + b)/5; subtract the two equations to get x - y = -a - 3b. So, x2 – y2 = (x + y)(x – y) = (9a + b)/5 (–a – 3b) = (–\[(9a)\]2 – 28ab – \[(3b)\]2)/5

The best way to solve this problem is to plug in an appropriate value for a. For example, plug-in 11 for a because 11 divided by 7 will give us a remainder of 4.

Then 3_a + 5_, where a = 11, gives us 38. Then 38 divided by 3 gives a remainder of 2.

The algebra method is as follows:

a divided by 7 gives us some positive integer b, with a remainder of 4.

Thus,

a / 7 = b 4/7

a / 7 = (7_b +_ 4) / 7

a = (7_b_ + 4)

then 3_a + 5 =_ 3 (7_b_ + 4) + 5

(3_a_+5)/3 = \[3(7_b_ + 4) + 5\] / 3

= (7_b_ + 4) + 5/3

The first half of this expression (7_b_ + 4) is a positive integer, but the second half of this expression (5/3) gives us a remainder of 2.

Divide the coefficients and subtract the exponents.

You need to factor to find the possible values for x. You need to fill in the blanks with two numbers with a sum of -12 and a product of 36. In both sets of parenthesis, you know you will be subtracting since a negative times a negative is a positive and a negative plus a negative is a negative

(x –__)(x –__).

You should realize that 6 fits into both blanks.

You must now set each set of parenthesis equal to 0.

x – 6 = 0; x – 6 = 0

Solve both equations: x = 6