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Mamont248 [21]
3 years ago
13

Helps me solve this problem please

Mathematics
1 answer:
____ [38]3 years ago
5 0

Step-by-step explanation:

2x + 5y = 6

move it = y= -2x+6

the slope is -2

so, the slope is negative

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Please answer this hurry up thank you so much!
Pepsi [2]
15 liters of water for 40 flower pots
3 0
3 years ago
-2/3- (-3/4)<br><br> Simplify if needed <br> Brainliest <br><br> Also, 5/6+5/12
bazaltina [42]

Answer:

-\frac{2}{3} - (-\frac{3}{4}) = \frac{1}{12}

\frac{5}{6} + \frac{5}{12} = \frac{5}{3}

Step-by-step explanation:

Given

-\frac{2}{3} - (-\frac{3}{4})

Required

Solve:

-\frac{2}{3} - (-\frac{3}{4})

Open bracket

-\frac{2}{3} - (-\frac{3}{4}) = -\frac{2}{3} +\frac{3}{4}

Take LCM

-\frac{2}{3} - (-\frac{3}{4}) = \frac{-8+9}{12}

-\frac{2}{3} - (-\frac{3}{4}) = \frac{1}{12}

\frac{5}{6} + \frac{5}{12}

Take LCM

\frac{5}{6} + \frac{5}{12} = \frac{10+5}{12}

\frac{5}{6} + \frac{5}{12} = \frac{15}{12}

Divide by 3/3

\frac{5}{6} + \frac{5}{12} = \frac{5}{3}

4 0
3 years ago
What is the range of the graph shown below
RUDIKE [14]
Theres nothing there tho
8 0
3 years ago
The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section
lisabon 2012 [21]

Answer:

\therefore y_2(x)=-\frac{e^{-6x}}{8}

The general solution is

y=c_1e^{2x}-c_2.\frac{e^{-6x}}{8}

Step-by-step explanation:

Given differential equation is

y''-4y'+4y=0

and y_1(x)=e^{2x}

To find the y_2(x) we are applying the following formula,

y_2(x)=y_1(x)\int \frac{e^{-\int P(x) dx}}{y_1^2(x)} \ dx

The general form of equation is

y''+P(x)y'+Q(x)y=0

Comparing the general form of the differential equation to the given differential equation,

So, P(x)= - 4

\therefore y_2(x)=e^{2x}\int \frac{e^{-\int 4dx}}{(e^{2x})^2}dx

           =e^{2x}\int \frac{e^{-4x}}{e^{4x}}dx

           =e^{2x}\int e^{-4x-4x} \ dx

            =e^{2x}\int e^{-8x} \ dx

            =e^{2x}. \frac{e^{-8x}}{-8}

           =-\frac{e^{-6x}}{8}

\therefore y_2(x)=-\frac{e^{-6x}}{8}

The general solution is

y=c_1e^{2x}-c_2.\frac{e^{-6x}}{8}

4 0
3 years ago
Write an equation of the line perpendicular to
AysviL [449]

Answer:

From general equation of a line:

y = mx + b

m » slope, b » y-intercept.

but perpendicularly:

m _{1} \times m _{2} =  - 1 \\  - 3 \times m =  - 1 \\ m =  \frac{1}{3}

at point (6, 2):

2 = ( \frac{1}{3}  \times 6) + b \\ 2 = 2 + b \\ b = 0

therefore, equation is :

{ \boxed{ \boxed{y =  \frac{1}{3} x}}}

8 0
3 years ago
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