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AleksandrR [38]
4 years ago
6

What is the average rate of change for this function for the interval fron x=2 to x=4

Mathematics
1 answer:
mezya [45]4 years ago
6 0
Subtract the two Y values of X = 2 and X = 4 to find the total difference, then divide by the difference between X=2 and X =4

When X = 2 Y = 9
When X = 4 Y = 64

Difference between the two Y values = 64 -9 = 55
Difference between the two X values = 4-2 = 2

The rate of change = difference in Y divided by difference in X:

55/2 = 27.5
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Element X is a radioactive isotope such that every 51 years, its mass decreases by half. Given that the initial mass of a sample
pychu [463]

Answer: It will take 6 years for 7500 grams of X to reach 6900 grams.

Step-by-step explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

a) for completion of half life:  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

51=\frac{0.69}{k}

k=\frac{0.69}{51}=0.0135years^{-1}

b) for  7500 g to reach to 6900 g

t=\frac{2.303}{0.0135}\log\frac{7500}{6900}

t=6years

It will take 6 years for 7500 grams of X to reach 6900 grams.

5 0
3 years ago
Find the ratio of : 800 ml to 4.8 litres​
Ray Of Light [21]

Answer:

The ratio is \frac{1}{6}

Step-by-step explanation:

We have

\frac{800mL}{4.8L} \\\\\frac{0.8L}{4.8L}\\\\\frac{1}{6}

5 0
2 years ago
Solve for x.each figure is a parallelogram.show steps
Sav [38]

Answer:

opposite sides of a parallelogram are equal

2x+15=x+15

2x-x=15-15

x=0

8 0
3 years ago
Read 2 more answers
Help asap worth 50 points<br> Use grouping symbols to make each equation true.<br> 16-4/2+3=9
ira [324]

Answer:

This question is only if 25 points. And you have typed 50 points.

Answer is 16-4-2-1

5 0
2 years ago
The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard dev
prohojiy [21]
<span>The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50. what is the probability that a student uses more than 580 minutes?

Given
&mu;=500
&sigma;=50
X=580
P(x<X)=Z((580-500)/50)=Z(1.6)=0.9452
=>
P(x>X)=1-P(x<X)=1-0.9452=0.0548=5.48%

</span>
7 0
3 years ago
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