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xxTIMURxx [149]

3 years ago

10

Lora read 1/8 of a comic book in 2 days she read the same number of pages everyday what part of the book does she read in a week

?

1 answer:

kipiarov [429]3 years ago

8 0

She reads 1/16 everyday. So there's 7 days in a week. So Lora reads 7/16 of the book in a week.

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Please help the problem is in the picture provided below.

Nadya [2.5K]

Answer:

sorry I don't know

Step-by-step explanation:

4 0

2 years ago

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What are the answers for 2, 3 , and 4? That’s it

PIT_PIT [208]

Answer:

1. Formula: b * h ÷ 2

12 * 3 = 36 ÷ 2 = 18 in^2

2. Formula: b * h ÷ 2

5 * 8 = 40 ÷ 2 = 20 ft^2

I could only help on two. Hope it helps though! =)

5 0

2 years ago

Solve the system of equations by graphing, elimination or

vladimir1956 [14]

Answer:

Solve the system of equations by graphing, elimination or

subod

Step-by-step explanation:

8 0

2 years ago

VERY EASY, WILL GIVE 50 POINTS FOR CORRECT ANSWER ASAP AND WILL GIVE BRAINLIEST.

mario62 [17]

Answer:

Coordinates switch, and the new y-coordinate changes sign

Explanation:

We have point A with the coordinates (-1,2), when we rotate it c90º it becomes (2,1).

What happened? The coordinates switched and the x-coordinate of A (-1) became the y-coordinate of A' (1) and it changed its sign.

7 0

3 years ago

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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