Answer:
The player throws 127.3 ft from second base to home plate.
Step-by-step explanation:
Given:
Distance from home to first base = 90 ft
Distance from first base to second base = 90 ft
We need to find the distance from second base to home.
Solution:
Now we can assume the complete scenario to be formed as a right angled triangle with two sides given and to find the third side.
Now by using Pythagoras theorem which states that;
Square of the hypotenuse side is equal to sum of squares of other two sides.
framing in equation form we get;
distance from second base to home = 
Rounding to nearest tent we get;
distance from second base to home = 127.3 ft
Hence The player throws 127.3 ft from second base to home plate.
If the perimeter is 44 the the diagonal is 22
The answer would be b A≈1636.8
Let P = number of coins of pennies (1 penny = 1 cent)
Let N = number of coins of nickels (1 nickel = 5 cents)
Let D = number of coins of dimes (1 dime = 10 cents)
Let Q = number of coins of quarters (1 quarter = 25 cents)
a) P + N + D + Q = 284 coins, but P = 173 coins, then:
173 + N + D + Q =284 coins
(1) N + D + Q = 111 coins
b) D = N + 5 OR D - N =5 coins
(2) D - N = 5 coins
c) Let's find the VALUE in CENTS of (1) that is N + D + Q = 111 coins
5N + 10D + 25 Q = 2,278 - 173 (1 PENNY)
(3) 5N + 10D + 25Q = 2105 cents
Now we have 3 equation with 3 variables:
(1) N + D + Q = 111 coins
(2) D - N = 5 coins
(3) 5N + 10D + 25Q = 2105 cents
Solving it gives:
17 coins N ( x 5 = 85 cents)
22 coins D ( x 10 = 220 cents)
72 coins D ( x 25 = 1,800 cents)
and 173 P,
proof:
that makes a total of 85+2201800+172 =2,278 c or $22.78
