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Vlad [161]
3 years ago
7

What is power rule and definitions

Mathematics
1 answer:
Grace [21]3 years ago
6 0
The power rule is one of several rules used to calculate the derivative of a function

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Please help me i’m failing
tatiyna
A. 2x - 3y is the answer
3 0
3 years ago
Which number should.be added to the data so that the range of the data is 31? 54,72,64,57
Sergio039 [100]

Answer:  41 or 85

<u>Step-by-step explanation:</u>

Range is the difference between the largest number and the smallest number.

The largest number is 72.  72 - 31 = 41

The smallest number is 54.  54 + 31 = 85

Adding either ONE of those numbers will result in a range of 31.

3 0
3 years ago
I need help, please.
Usimov [2.4K]
2 rooms would be 62+42 which is 104

3 rooms would be 104+42 which is 146

4 rooms would be 146+42 which is 188

5 rooms would be 188+42 which is 230

6 rooms would be 230+42 which is 272

but these are the prices including the 20$

without the 20$ the prices would be

2 rooms is 84

3 rooms is 126

4 rooms is 168

5 rooms is 210

6 rooms is 252
4 0
2 years ago
Read 2 more answers
Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2
igor_vitrenko [27]

Answer:  The required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Step-by-step explanation:  We are given to find the solution of the following initial value problem :

y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.

Let y=e^{mx} be an auxiliary solution of the given differential equation.

Then, we have

y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.

Substituting these values in the given differential equation, we have

m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

The given conditions implies that

y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.

From equation (i), we get

\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.

Substituting the values of A and B in the general solution, we get

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Thus, the required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

4 0
3 years ago
A coin is tossed 8 times. What is the probability of getting all heads? Express your answer as a simplified fraction or a decima
kumpel [21]

Answer:

The probability of getting all heads is \frac{1}{256}.

Step-by-step explanation:

For each time the coin is tossed, there are only two possible outcomes. Either it is heads, or it is not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem, we have that:

A coin is tossed 8 times. This means that n = 8

In each coin toss, heads or tails are equally as likely. So p = \frac{1}{2}

What is the probability of getting all heads?

This is P(X = 8)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

[tex]P(X = 8) = C_{8,8}*(\frac{1}{2})^{8}*(1 - \frac{1}{2})^{0} = \frac{1}{256}

The probability of getting all heads is \frac{1}{256}.

7 0
3 years ago
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