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FinnZ [79.3K]
2 years ago
14

What property is 7+(5+b)=(5+b)+7

Mathematics
1 answer:
denis23 [38]2 years ago
3 0

Answer:

The answer is zero or cannot be solved

Step-by-step explanation:

According to the question given,we will have to deal with the bracket first

7+(5+b)=(5+b)+7

7+5+b=5+b+7

12+b=12+b

There is no way to solve this question further because any further solution will definitely give us zero

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the distance

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More time care travels, more distance it goes. So distance depends of the time.

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natulia [17]

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C

Step-by-step explanation:

In a Pythagorean triple the square on the longest side is equal to the sum of the squares on the other 2 sides.

A

longest side = 10 ⇒ 10² = 100

6² + 8² = 36 + 64 = 100

A is a Pythagorean triple

B

longest side = 17 ⇒ 17² = 289

8² + 15² = 64 + 225 = 289

B is a Pythagorean triple

C

longest side = 9 ⇒ 9² = 81

5² + 8² = 25 + 64 = 89

89 ≠ 81

C is not a Pythagorean triple

D

longest side = 13 ⇒ 13² = 169

5² + 12² = 25 + 144 = 169

D is a Pythagorean triple

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2 years ago
True or false?
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Step-by-step explanation:

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3 years ago
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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
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