Answer:
7*10*10*10 = 7000
7*10 = 70
70*10 = 700
700*10 = 7000
Step-by-step explanation:
The given expression is:
7*10^3
Here 10^3 means that 10 will be multiplied 3 times:
7*10*10*10 = 7000
How did we get 7000?
7*10*10*10 = 7000
7*10 = 70
70*10 = 700
700*10 = 7000
We can also say that there are 7 1000s in 7000....
Determine whether the relation is a function. {(−3,−6),(−2,−4),(−1,−2),(0,0),(1,2),(2,4),(3,6)}
Gennadij [26K]
Answer:
The relation is a function.
Step-by-step explanation:
In order for the relation to be a function, every input must only have one output. Basically, you can't have 2 outputs for 1 input but you can have 2 inputs for 1 output. Looking at all of the points in the relation, we see that no input has multiple outputs, so the answer is yes, the relation is a function.
Answer:
160% of 35 is 56.
Step-by-step explanation:

y · 160 = 56 · 100
160y = 5600
160y ÷ 160 = 5600 ÷ 160
y = 35
Answer:
The correct answer is:
Between 600 and 700 years (B)
Step-by-step explanation:
At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

First, let us calculate the decay constant (k)

Next, let us calculate the half-life as follows:

Therefore the half-life is between 600 and 700 years