Question

problems 4 - 5: here is a table representing a pattern.

1. circle the equation that represents the table.

a. $y = 60+\frac{1}{2}x$
b. $y = 60cdot(\frac{1}{2})^x$
c. $y = 60cdot2^x$
d. $y = 60-\frac{1}{2}x$

1. explain your thinking.

Explanation:

Step1: Test x = 0

When \(x = 0\), substitute into each option.

• Option A: \(y=60+\frac{1}{2}\times0 = 60\)
• Option B: \(y = 60\times(\frac{1}{2})^0=60\times1 = 60\)
• Option C: \(y = 60\times2^0=60\times1 = 60\)
• Option D: \(y = 60-\frac{1}{2}\times0 = 60\)

All options match when \(x = 0\).

Step2: Test x = 1

When \(x = 1\), substitute into each option.

• Option A: \(y=60+\frac{1}{2}\times1=60 + 0.5=60.5

eq30\)

• Option B: \(y = 60\times(\frac{1}{2})^1=60\times\frac{1}{2}=30\)
• Option C: \(y = 60\times2^1=120

eq30\)

• Option D: \(y = 60-\frac{1}{2}\times1=60 - 0.5 = 59.5

eq30\)
Only option B matches when \(x = 1\).

Step3: Test x = 2 for confirmation

When \(x = 2\), for option B: \(y = 60\times(\frac{1}{2})^2=60\times\frac{1}{4}=15\) which is consistent with the table.

for 5:
We tested the values of \(x\) from the table in each of the given equations. When \(x = 0\), all equations gave \(y = 60\). But when \(x = 1\), only the equation \(y = 60\cdot(\frac{1}{2})^x\) gave \(y = 30\) as in the table. Further, when \(x = 2\), substituting into \(y = 60\cdot(\frac{1}{2})^x\) gave \(y = 15\) which also matched the table. So the equation \(y = 60\cdot(\frac{1}{2})^x\) represents the pattern in the table.

Answer:

B. \(y = 60\cdot(\frac{1}{2})^x\)