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3 years ago

We want to change .8333 to a fraction

Mathematics

1 answer:

OleMash [197]3 years ago

4 0

Let x =0.8333.... Then 10x=8.3333-----(1) 100x=83.333..----(2) So (2)-(1) ,90x=75 x=75/90 =7.5/9 Hope it helps !!!!

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Luisa recieved $108 for a necklace she sold on an online auction. The auction site took a fe of 10% of selling price. What was t

Vera_Pavlovna [14]

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2 years ago

An equilateral triangle has side length 10 centimeters. Find its area in square centimeters. HINT: The altitude of an equilatera

dmitriy555 [2]

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3 years ago

Solve the system by graphing. Y=2x-9 y=-2x-1

sattari [20]

Answer:

(2,-5)

Step-by-step explanation:

See attachment

One can also solve this by calculation:

y=2x-9

y=-2x-1

Rearrange either equation to find x. I'll use the first:

y=2x-9

2x = y+9

x = (y+9)/2

Now use this value of x in the second equation:

y = -2x-1

y =-2((y+9)/2)-1

y = (-2y-18)/2)-1

y = -y -9 - 1

2y = -10

y = -5

Now use -5 for y in the rearranged equation:

y = -2x-1

-5 = -2x-1

-2x = -4

x = 2

Solution is (2,-5)

But the question wants a graph solution, which is also fun when you use DESMOS.

4 0

2 years ago

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:

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

7 0

3 years ago

How do I solve this problem?

Morgarella [4.7K]

Next time use the desmos calculator online and input the numbers.

The answer to this equation is:

y = (x - 3)^2 - 2

5 0

3 years ago