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2 years ago

Lim x tends to 0 sinax/tanbx

1 answer:

8 0

Express tan in terms of sin and cos

= sin ax cos bx / sin bx

Now I'm stuck! sorry

For what its worth Ive drawn the graph and the answer is a/b

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Carbon-141414 is an element that loses about 10\%10%10, percent of its mass every millennium (i.e., 100010001000 years). A sampl

ololo11 [35]

Answer:

Carbon-14 loses around 10% ( 0.1 in decimal form) of its mass, after one millennium.

Then if we start with a mass A of Carbon-14, after one millennium we will have a mass equal to:

A - A*0.1 = A*(0.9)

After another millennium, we will have a mass equal to:

A*(0.9) - A*(0.9)*0.1 = A*(0.9)^2

And so on, this is an exponential decay.

We already can see the pattern here, after x millenniums, the mass of carbon-14 will be:

M(x) = A*(0.9)^x

Now, in this problem we have 600 grams of carbon-14, then the equation for the mass will be:

y = M(x) = 600g*(0.9)^x

And the graph of this equation is shown below.

4 0

2 years ago

Read 2 more answers

What is the product and simplest form 2/3 times 1/8

ludmilkaskok [199]

Answer:

$\frac{1}{12}$

Step-by-step explanation:

Since we're finding the product, we have to multiply:

$\frac{2}{3}$ × $\frac{1}{8}$

You can simplify in this stage by using the "butterfly method", and dividing $2$ by $2$ , and $8$ by $2$ , you'd then have:

$\frac{1}{3}$ × $\frac{1}{4}$

Multiply the numerators and the denominators to get:

$\frac{1}{12}$

If you prefer the longer way, again, multiply:

$\frac{2}{3}$ × $\frac{1}{8}$

Multiply the numerators and the denominators:

$\frac{2}{24}$

Simplify the fraction by dividing both the numerator and denominator by $2$ :

$\frac{1}{12}$

7 0

2 years ago

Read 2 more answers

If 8 identical blackboards are to be divided among 4 schools,how many divisions are possible? How many, if each school mustrecei

MAXImum [283]

Answer:

There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.

Step-by-step explanation:

Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.

The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is ${11 \choose 3} = 165 .$ As a result, we have 165 ways to distribute the blackboards.

If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is ${7 \choose 3} = 35$ . Thus, there are only 35 ways to distribute the blackboards in this case.