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2 years ago

Order of operations-I'm not sure what the order is, please help.

Mathematics

2 answers:

Nostrana [21]2 years ago

6 0

Answer:

SO..........basically Order of operations are very easy. In a problem like this

(number 2) 25-14+3

The first thing you do would be to addition

The order of operation goes PEMDAS

P= parenthesis

E= Exponents

M= Mutpilcation

D= Divison

A=addition

S=Subtraction

so it would be 25-14+3

\ /

25-17

Then it would be 25-17

\ /

This is the answer to number 2 I hope it answers your question about the problem and what order of operation is

P.S How I remember PEMDAS is Please

Excuse

Dear

Aunt

Sally

Step-by-step explanation:

LuckyWell [14K]2 years ago

3 0

Answer:

Order of operations is and order that you go by addition, subtraction, multiplication, division, exponentiation, and grouping.

Usually its just the first four because that's usually what you get in home work or assignments. If there is no addition you skip to multiplication and so forth. And the equation in parentheses usually goes first before you solve anything else. .

For example in your first question you would do addition first then subtraction. Since there is no multiplication or division you don't have to include that. 34 + 17 = 51 - 5 = 46

*eliza*

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zvonat [6]

Answer:

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2 years ago

Read 2 more answers

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

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2 years ago

Is 2/5 - (-5/6) positive,negative, or zero

malfutka [58]

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2 years ago

Order of operations-I'm not sure what the order is, please help.​

Order of operations-I'm not sure what the order is, please help.