25x+20y=455
x+y=20
20x+20y=400
-25x-20y=-455
-5x=-55
x=11
11+y=20
y=9
Solved it as well, sorry if I wasn't supposed to
The value of the population of the growth of an endangered birth after 5 years is 1975
<h3>How to determine the population after 5 years?</h3>
The population function is given as:
B(t) = 100 + 3/5t^5
At 5 years, the value of t is 5
So, we have
t = 5
Next, we substitute 5 for t in the equation B(t) = 100 + 3/5t^5
This gives
B(5) = 100 + 3/5 * 5^5
Evaluate the exponent
B(5) = 100 + 3/5 * 3125
Evaluate the product
B(5) = 100 + 1875
Evaluate the sum
B(5) = 1975
Hence, the value of the population of the growth of an endangered birth after 5 years is 1975
Read more about exponential functions at:
brainly.com/question/2456547
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Answer:
FG = 44
EG = 80
Step-by-step explanation:
For FG:
1) 5z - 16 = 3z + 8
2) 5z = 3z + 24
3) 2z = 24
4) z = 12
5) 3(12) + 8 = 44
For EG:
1) 2w + 22 = 4w + 4
2) 2w = 4w - 18
3) -2w = -18
4) w = 9
5) 4(9) + 4 + 2w + 22 = 80
Answer:
Step-by-step explanation:
Given that:
Distance traveled by seal to catch the fish = 15 below sea level
Sea lion wanted to catch a larger fish so sea lion dove 6 feet lesser than two time the distance that the seal traveled.
To find:
The expression to represent the position of the sea lion in the sea.
Solution:
If the distance traveled by seal is represented by 
then as per the question statement:
Twice of = 
6 lesser than twice of = - 6
Now, putting the value of = 15
Therefore, the expression to represent the sea lion's position w.r.to sea level.
