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artcher [175]
3 years ago
14

What is the slope of line that is parallel to y=2x+1

Mathematics
1 answer:
schepotkina [342]3 years ago
7 0

Answer:

2x

Step-by-step explanation:

Parallel lines have the same gradient

Therefore it would have the same gradient of 2x

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Please i really need help on this.
Studentka2010 [4]

Answer:

it is C

Step-by-step explanation: please brainliest

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2 years ago
You drive your car for three hours at an average speed of 130 km/h how far did you go
aliina [53]
About 390 km because you drive 130 km per hour so 130 times 3 is 390
8 0
3 years ago
Read 2 more answers
Loretta grew 54 watermelons in her garden last year. This year, that number increased by 5/9. Which expression shows the number
Mars2501 [29]

Answer:

54+\frac{5}{9}\times 54

Step-by-step explanation:

We are given that

Loretta grew  watermelons in her garden last year=54

We have to find the expression which shows the number of watermelons Loretta grew this year.

To find the expression which shows the  number of watermelons Loretta grew this year by adding watermelons grew in her garden last year and number of watermelons increased .

According to question

5/9 of 54=\frac{5}{9}\times 54

Therefore, the expression which shows the  number of watermelons Loretta grew this year

=54+\frac{5}{9}\times 54

6 0
2 years ago
1/8 + c = 4/5 <br><br>what is c
ASHA 777 [7]

Answer:

27/40

Step-by-step explanation:

7 0
3 years ago
Find a particular solution to y" - y + y = 2 sin(3x)
leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

     =e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}

Multiplying both sides of equation (1) by integrating factor and integrating we get

\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I

I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx

8 0
2 years ago
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