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steposvetlana [31]
3 years ago
6

Please help i don’t know this

Mathematics
2 answers:
AnnZ [28]3 years ago
3 0
-1,1x + 6,4 > -1,3 <=>

-1,1x > - 7,7 <=>

x < 7,7/1,1 <=>

x < 7

The answer is the third graph.
gayaneshka [121]3 years ago
3 0

Answer:

it is the third option

Step-by-step explanation:

you would subtract the 6.4 to the other side of the equation and then you would divide it by -1.1 and you would get 7

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Does anyone know how to do number 8?
Alekssandra [29.7K]

\frac{a - 1}{7 - 7a}  =  \frac{a - 1}{7(1 - a)}  =  \frac{ - (1 - a)}{7(1 - a)}  =  \frac{ - 1}{7}
for all a except a=1
which is choice 3

8 0
4 years ago
8.
Naya [18.7K]

Answer:

63.6cm²

Step-by-step explanation:

Area of square = 81

Side of square = \sqrt{81}

                        = 9

-----------------------------------------

diameter of circle = 9

radius of circle = 9 ÷ 2

                        = 4.5

area of circle = πr²

                      = π (4.5)²

                      = 63.6172512  

                      ≈ 63.6 cm²

3 0
3 years ago
1. Use separation of variables to find the solution to the differential equation subject to the given initial condition.
andrew11 [14]

Answer:

Step-by-step explanation:

Given the differential equation dy/dx = 5y/x subject to the condition y = 4 and x = 1. Using the variable separable method of solving differential equation, we will have;

dy/dx = 5y/x

Separate the variables

dy/5y = dx/x

Integrate both sides of the expression

\frac{1}{5}\int\limits \frac{1}{y}  \, dy = \int\limits \frac{dx}{x} \\ \\\frac{1}{5}lny = lnx + C\\\\lny = 5lnx+5C\\

using the initial condition y = 4 while x = 1

ln4 = 5ln1 + 5C

ln4 = 0+5C

C = ln4/5

Substituting the value of C back into the expression;

lny = 5 lnx+5(ln4/5)\\lny = 5lnx+ln4\\lny = lnx^5+ln4\\lny = ln(4x^5)\\y = 4x^5

<em>Hence the solution to the differential equation is y = 4x⁵</em>

<em></em>

b) Given 4(du/dt) = u²

du/dt = u²/4

du/ u² = dt/4

u⁻²du = 1/4 dt

integrate both sides of the equation

\int\limit {u^{-2}} \, du  = \int\limits\frac{1}{4}  \, dt\\\\\frac{u^{-1}}{-1} = \frac{t}{4} + C\\\\\frac{-1}{u} =  \frac{t}{4} + C

Imputing the initial condition u(0) = 7 i.e when t = 0, u = 7

\frac{-1}{7} =  \frac{0}{4} + C\\\\\frac{-1}{7} =  C\\

\frac{-1}{u} =  \frac{t}{4} - \frac{1}{7}

<em>Hence the solution to the DE is </em>\frac{-1}{u} =  \frac{t}{4} - \frac{1}{7}

6 0
3 years ago
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