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oee [108]
3 years ago
15

SOMEBODY PLEASE HELP ME!!!!!!!!!!!!! PLEASE!!!!! I WILL AWARD BRAINLIEST!!!!!!!

Mathematics
1 answer:
seraphim [82]3 years ago
8 0

Answer:

1. 18

2. 32

Step-by-step explanation:

1.  xy  is the digit

Remember x is the 10 digit so we multiply it by 10

2(x+y) = x *10 +y

Distribute

2x+2y = 10x+y

Subtract y from each side

2x +2y -y = 10x +y - y

2x+y = 10x

Subtract 2x from each side

2x+y -2x = 10x-2x

y = 8x

Since these are digits x must be 1 or y would be bigger than a digit

Then y =8

Our two digit number is 18

2.  1.  xy  is the digit

Remember x is the 10 digit so we multiply it by 10

10x +y =xy+26

Subtract xy from each side

10x -xy + y = 26

Factor out an -y

10x -y(x-1) = 26

X must be bigger than 2 or we cannot get 26

Let x=3

30 -y(3-1) =26

30 -2y = 26

Subtract 30 from each side

-2y = -4

Divide by -2

y=2

The number is 32

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Section 5.2 Problem 21:
Fittoniya [83]

Answer:

y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)] (See attached graph)

Step-by-step explanation:

To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation am^2+bm+c=0 where the values of m are the roots:

y''+4y'+10y=0\\\\m^2+4m+10=0\\\\m^2+4m+10-6=0-6\\\\m^2+4m+4=-6\\\\(m+2)^2=-6\\\\m+2=\pm\sqrt{6}i\\\\m=-2\pm\sqrt{6}i

Since the values of m are complex conjugate roots, then the general solution is y(x)=e^{\alpha x}[C_1cos(\beta x)+C_2sin(\beta x)] where m=\alpha\pm\beta i.

Thus, the general solution for our given differential equation is y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)].

To account for both initial conditions, take the derivative of y(x), thus, y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]

Now, we can create our system of equations given our initial conditions:

y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(0)=e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]=3\\\\C_1=3

y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]\\\\y'(0)=-2e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]+e^{-2(0)}[-C_1\sqrt{6}sin(\sqrt{6}(0))+C_2\sqrt{6}cos(\sqrt{6}(0))]=-2\\\\-2C_1+\sqrt{6}C_2=-2

We then solve the system of equations, which becomes easy since we already know that C_1=3:

-2C_1+\sqrt{6}C_2=-2\\\\-2(3)+\sqrt{6}C_2=-2\\\\-6+\sqrt{6}C_2=-2\\\\\sqrt{6}C_2=4\\\\C_2=\frac{4}{\sqrt{6}}\\ \\C_2=\frac{4\sqrt{6}}{6}\\ \\C_2=\frac{2\sqrt{6}}{3}

Thus, our final solution is:

y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]

7 0
2 years ago
Whitney works at an aquarium. She uses hoses of different sizes to fill the fish tanks. Hose A takes 16 min to fill a 160-gal ta
scZoUnD [109]

Answer:

Hose A fills at a rate of 10 gal per min

Hose B fills at a rate of 8 gal per min

Hose B fills at a faster rate

Step-by-step explanation:

7 0
3 years ago
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I need help with this question
Molodets [167]
1. You have that Chris used 18 gallons of gas in 420 miles.

 2. Therefore: If Chris drove 420 miles using 18 gallons of gas, how many  gallons would in 357 miles?

 3. Let's call "x" to the amount of gallons he need to drive 357 miles. So, you have:

 420 miles----18 gallons
 357 miles-----x

 x=(357)(18)/420
 x=6426/420
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4 0
3 years ago
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choli [55]
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3 years ago
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If AABC ~ ADEF, what is the length of DE?<br><br> NEED HELP ASAP please and thanks In advance
chubhunter [2.5K]

Answer:

DE = 21

Step-by-step explanation:

Recall: the ratio of the corresponding side lengths of similar triangles are equal

Given that ∆ABC ~ ∆DEF, therefore,

AB/DE = BC/EF = AC/DF

AB = 8.4

DE = x

BC = 10

EF = 25

AC = 16.5

DF = 41.25

Let's find DE using AB/DE = BC/EF. Thus:

8.4/x = 10/25

Cross multiply

x*10 = 25*8.4

10x = 210

Divide both sides by 10

x = 210/10

x = 21

DE = x = 21

5 0
3 years ago
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