Answer:324
Step-by-step explanation:
int i = 42.7; /* konwersja z double do int */
float f = i; /* konwersja z int do float */
double d = f; /* konwersja z float do double */
unsigned u = i; /* konwersja z int do unsigned int */
f = 4.2; /* konwersja z double do float */
i = d; /* konwersja z double do int */
char *str = "foo"; /* konwersja z const char* do char* [1] */
const char *cstr = str; /* konwersja z char* do const char* */
void *ptr = str; /* konwersja z char* do void* */
Podcza
2000000 + 900000 + 30000 + 7000 + 80 + 2
The largest possible volume of the given box is; 96.28 ft³
<h3>How to maximize volume of a box?</h3>
Let b be the length and the width of the base (length and width are the same since the base is square).
Let h be the height of the box.
The surface area of the box is;
S = b² + 4bh
We are given S = 100 ft². Thus;
b² + 4bh = 100
h = (100 - b²)/4b
Volume of the box in terms of b will be;
V(b) = b²h = b² * (100 - b²)/4b
V(b) = 25b - b³/4
The volume is maximum when dV/db = 0. Thus;
dV/db = 25 - 3b²/4
25 - 3b²/4 = 0
√(100/3) = b
b = 5.77 ft
Thus;
h = (100 - (√(100/3)²)/4(5.77)
h = 2.8885 ft
Thus;
Largest volume = [√(100/3)]² * 2.8885
Largest Volume = 96.28 ft³
Read more about Maximizing Volume at; brainly.com/question/1869299
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Answer:
X=82
Step-by-step explanation:
Answer:
1 17/24
Step-by-step explanation:
