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15 days ago

5

Pete's dad usually deposits $25 into his bank account every week. for the next 8 weeks , he wants to deposit more than the usual

$25, and he wants the extra amount deposited to be the same each week. which inequality and solution show how much more he can deposit each week and keep the total of his deposit above 500 for those 8 weeks? 8(25+x)>500;x>$300
25+8x>500;x>$59.37
8(25+x)>500;x>37.50
8(25)+x>500;x>$300

2 answers:

Svet_ta [4.3K]15 days ago

8 0

Option B is correct, expressed as 8(25 + x) > 500; solving for x gives x > 37.50

Svet_ta [4.3K]15 days ago

4 0

Answer:

$8(25+x)>500;x>37.5$

Detailed Solution:

Given: He plans to deposit more than his regular $25 weekly amount over 8 weeks.

Find: The inequality and solution representing the consistent extra amount he can add weekly so that his total savings after 8 weeks exceed $500.

Approach:

His regular weekly deposit is $25.

Define x as the constant additional amount deposited each week.

Therefore, weekly deposit equals 25 plus x.

Over 8 weeks, total deposits amount to 8 times (25 plus x).

The total must be greater than 500.

Thus, $8(25+x)>500$

$200+8x>500$

$8x>300$

$x>\frac{300}{8}$

$x>37.5$

Inequality is: $8(25+x)>500$

Solution is: $x>37.5$

Therefore, choice C is the correct answer.

Hence, the inequality and solution expressing the additional weekly deposit to keep the total over 500 in 8 weeks is $8(25+x)>500;x>37.5$

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