The output would depend on the input so relating this to the question, the input is the time (in minute) and the output is the amount of water left in the tank
We can give a letter 't' for the time, the input, and f(t) for the output
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The original amount of water in the tank is 10450, so this will be the fixed constant.
The amount of water lost per minute is 270 so this will be the term that varies depends on the variable of time, we write this as 270t
The function is given f(t) = 10450 - 270t
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Given, t = 10
f(10) = 10450 - 270(10)
f(10) = 7750 ml
Greater than 90° but less than 180°.
I am sorry if i am wrong but i keep getting 6,4
Answer:
a) 1/27
b) 16
c) 1/8
Step-by-step explanation:
a) 
One of the properties of the exponents tells us that when we have a negative exponent we can express it in terms of its positive exponent by turning it into the denominator (and changing its sign), so we would have:

And now, solving for x = 9 we have:

b) 
This is already a positive rational exponent so we are just going to substitute the value of y = 8 into the expression

c) 
Using the same property we used in a) we have:

And now, solving for z = 16 we have:

It has to be A I’m sorry if I’m wrong