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Pepsi [2]
3 years ago
5

7+3√8 please help me

Mathematics
2 answers:
-BARSIC- [3]3 years ago
4 0
15.4/ 15.4852813742
madam [21]3 years ago
3 0

Answer:

look at the picture i have sent

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What is an equation of the line that passes through the points (2, -1) and<br> (3,-4)?
IRINA_888 [86]

Answer:

y=3x+5

Step-by-step explanation:

8 0
3 years ago
The product of two consecutive, negative integers is 182. What equation represents the situation?
Sholpan [36]
-13 and -14

They both are consecutive negative integers that multiply to 182.
4 0
3 years ago
Read 2 more answers
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 <em>c)</em>

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

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

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 <em>A</em> and <em>B</em> 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
What’s the greatest common factor of 12x^2+9x^3-6x^4
VikaD [51]

12x^2+9x^3-6x^4\\\\LCD(12,\ 9,\ 6)=3\\\\LCD(x^2,\ x^3,\ x^4)=x^2\\\\12x^2=(3x^2)(4)\\9x^3=(3x^2)(3x)\\-6x^4=(3x^2)(-2x^2)\\\\GCF\ of\ 12x^2+9x^3-6x^4\ is\ equal\ \boxed{3x^2}\\\\12x^2+9x^3-6x^4=(3x^2)(4+3x-2x^2)

7 0
3 years ago
Enter the range of values for x: 48° 2x-12 16 22 AB=AD​
taurus [48]

Answer:

its 6 I just took the assignment

6 0
3 years ago
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