Answer:
You can put this solution on YOUR website!
the inequality is 500 - 25x >= 200
this insures that he will have at least 200 at the end of the summer.
subtract 200 from both sides of that inequality and add 25x to both sides of that inequality to get 500 - 200 >= 25x
simplify to get 300 >= 25x
divide both sides of that equation by 25 to get 300 / 25 >= x
simplify to get 12 >= x
12 >= x means x <= 12.
when x is smaller than or equal to 12, he will be guaranteed to have at least 200 in the account at the end of the summer.
when x = 12, what is left in the account is 500 - 25 * 12 = 200.
when x = 11, what is left in the account is 500 - 25 * 11 = 225.
when x = 13, what is left in the account is 500 - 25 * 13 = 175.
the maximum number of weeks he can withdraw money from his account is 12.
Step-by-step explanation:
Answer:
rounding to the thousandths place, the answer is 0.714
Step-by-step explanation:
This can easily be looked up on google
If <em>x</em>² + <em>y</em>² = 1, then <em>y</em> = ±√(1 - <em>x</em>²).
Let <em>f(x)</em> = |<em>x</em>| + |±√(1 - <em>x</em>²)| = |<em>x</em>| + √(1 - <em>x</em>²).
If <em>x</em> < 0, we have |<em>x</em>| = -<em>x</em> ; otherwise, if <em>x</em> ≥ 0, then |<em>x</em>| = <em>x</em>.
• Case 1: suppose <em>x</em> < 0. Then
<em>f(x)</em> = -<em>x</em> + √(1 - <em>x</em>²)
<em>f'(x)</em> = -1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = -1/√2 → <em>y</em> = ±1/√2
• Case 2: suppose <em>x</em> ≥ 0. Then
<em>f(x)</em> = <em>x</em> + √(1 - <em>x</em>²)
<em>f'(x)</em> = 1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = 1/√2 → <em>y</em> = ±1/√2
In either case, |<em>x</em>| = |<em>y</em>| = 1/√2, so the maximum value of their sum is 2/√2 = √2.
.83
25 bus riding sophomores / 30 sophomores
Answer:
The circumference and area of a circle with a 17 ft diameter are:
C = 53.4 ft
A = 227 ft2
