First, simplify 3 ◊ 2: 4(3) − 2 = 12 − 2 = 10. Then, solve 10 ◊ 3: 4(10) − 3 = 40 − 3 = 37.
Plug the numbers in the grid into the formula, matching up the number in each section with the corresponding variable in the formula:
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Substitute 2B for G in the formula. Note that the term 2B appears in both the numerator and denominator, so they cancel out. |
Determine the maximum spring factor by setting x = 9.
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Pick numbers to see what happens to the cost when b is doubled. If the original value of b is 2, the cost is 16t. When b is doubled to 4, the new cost value is 256t. The cost has increased by a factor of
, or 16.
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The length of a side on the real sculpture is 4 m. |
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The length of a side on the model is 2 cm. |
V = s3 = (2)3 = 8 | The volume of the model is 8. |
Let c = competitive edge:
Pick numbers to see what happens to the competitive edge when W is tripled and L is halved. If the original value of W is 2 and the original value of L is 2, the original value of c is
.
If W triples to 6 and L is halved to 1, the new value of c is
. The competitive edge has increased from 1 to
.
Pick real numbers to solve this problem. Set the radius of the original circle equal to 2. Therefore, the radius of the new circle is equal to 6. Once you compute the areas of both circles, you can find the ratio:
An = 3 − 8n
A1 = 3 − 8(1) = 3 − 8 = −5
The easiest way to solve this problem is to write equations for S2 and S3 in terms of the other items in the sequence and solve for S2:
Now substitute the expression for S3 into the first equation and solve:
By this logic,
.
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k(3) = 27. Therefore:
First, find the output value of the inner function: g(4) = 16. Then, find
To find the values for which f(x) = g(x), set the functions equal to each other:
g(x) = |x − 1| − 1. This function is an absolute value, which typically has a V-shape. You can identify the correct graph by trying x = 0, which yields g(0) = 0, the origin. Then, try x = 1, which yields g(1) = −1 and the point (1, −1). Next, try x = 2: g(2) = |2 − 1| − 1 = 1 − 1 = 0. These three points fall on the V-shape.
The powers of 2 have a repeating pattern of four terms for their units digits: {2, 4, 8, 6}. That means that every fourth term, the pattern repeats. For instance, the fifth term has the same units as the first term, because 5 − 1 = 4. So terms that are four terms apart, or a multiple of four terms apart, will have the same units digit.
The 34th term and the 26th term are 34 − 26 = 8 terms apart. Because 8 is a multiple of 4, the terms will have the same units digit. The two quantities are equal. Incidentally, the units digit of A26 and A34 is 3.
P ■ Q = P + 2Q for all integers P and Q | ||
Quantity A | Quantity B | |
11 ■ 5 = (11) + 2 × (5) = 11 + 10 = 21 |
5 ■ 11 = (5) + 2 × (11) = 5 + 22 = 27 |
Quantity B is greater.
Plug in numbers to answer this question. Use a table to organize the information:
Old | New | |
---|---|---|
Length | 2 | 2 × 2 = 4 |
Width | 1 | W |
Area | 2 | 2 × 6 = 12 |
4 × W = 12
W = 3
Compare the new width to the original:
The width increased by a factor of 3.
The two quantities are equal.