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2 years ago

How many times can 14 go into 100

Mathematics

2 answers:

pogonyaev2 years ago

4 0

14 goes into 100 7.14285 repeating times.

morpeh [17]2 years ago

3 0

Answer:

$\frac{50}{7}$ times can 14 go into 100.

Step-by-step explanation:

Say x be the times when 14 can go into 100.

'Times' in the above statement means multiplication operation.

So, the equation for the given statement becomes:

$x\times 14=100$

To solve the value of 'x', we shift the number '14' to other side of the equation and operation applied will be division.

Solving the above equation, we get:

$x=\frac{100}{14}=\frac{50}{7}=7\frac{1}{7}$

$\frac{50}{7}$ times can 14 go into 100.

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Which of the following integrals cannot be evaluated using a simple substitution?

vitfil [10]

The third one, not sure about any other ones.

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2 years ago

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Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

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3 years ago

Simplify:

OLEGan [10]

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2 years ago

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Pls answer quick. Imma need it fast.

Tema [17]

Answer:

Step-by-step explanation:

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2 years ago

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Define a substitution method to solve the followingsysten of equation of y=2x-3, x+y=1

Ivan

I will solve you system by substitution

y = 2x - 3 ; x + y = 1

→Step 1: Solve y = 2x - 3 for y

→Step 2: Substitute 2x - 3 for y in x + y = 1:

x + y = 1

x + 2x- 3 = 1

3x - 3 = 1 (Simplify both sides of the equation)

3x - 3 + 3 = 1 + 3 (Add 3 both sides)

3x = 4

3x ÷ 3 = 4 ÷ 3 (Divide each side by 3)

x = 4/3

→Step 3: Substitute 4/3 for x in y = 2x - 3:

y = 2x - 3

y = 2 (4/3) -3

y = -1/3 (Simplify both sides of the equation)

Answer:

x = 4/3 and y = -1/3

∫Hope that helps∫

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2 years ago