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3 years ago

HELP!!! NEED ANSWER ASAP!

Mathematics

1 answer:

vitfil [10]3 years ago

8 0

Answer:

Yeah he's right...

Step-by-step explanation:

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Machine A can label 11 bottles in 1 minute. Machine B can label 14 bottles in 1 minute. How many bottles can both machines label

Andru [333]

Answer:

Machine 1: 165

Machine 2: 210

Step-by-step explanation:

Machine 1: 11 bottle per minute

11 * 15 = 165 labeled bottles per minute

Machine 2: 14 bottles per minute

14*15 = 210 labeled bottles per minute

7 0

3 years ago

Read 2 more answers

Solve the initial value problems.

slavikrds [6]

Both equations are linear, so I'll use the integrating factor method.

The first ODE

$xy' + (x+1)y = 0 \implies y' + \dfrac{x+1}x y = 0$

has integrating factor

$\exp\left(\displaystyle \int\frac{x+1}x \, dx\right) =\exp\left(x+\ln(x)\right) = xe^x$

In the original equation, multiply both sides by eˣ :

$xe^x y' + (x+1) e^x y = 0$

Observe that

d/dx [xeˣ] = eˣ + xeˣ = (x + 1) eˣ

so that the left side is the derivative of a product, namely

$\left(xe^xy\right)' = 0$

Integrate both sides with respect to x :

$\displaystyle \int \left(xe^xy\right)' \, dx = \int 0 \, dx$

$xe^xy = C$

Solve for y :

$y = \dfrac{C}{xe^x}$

Use the given initial condition to solve for C. When x = 1, y = 2, so

$2 = \dfrac{C}{1\cdot e^1} \implies C = 2e$

Then the particular solution is

$\boxed{y = \dfrac{2e}{xe^x} = \dfrac{2e^{1-x}}x}$

The second ODE

$(1+x^2)y' - 2xy = 0 \implies y' - \dfrac{2x}{1+x^2} y = 0$

has integrating factor

$\exp\left(\displaystyle \int -\frac{2x}{1+x^2} \, dx\right) = \exp\left(-\ln(1+x^2)\right) = \dfrac1{1+x^2}$

Multiply both sides of the equation by 1/(1 + x²) :

$\dfrac1{1+x^2} y' - \dfrac{2x}{(1+x^2)^2} y = 0$

and observe that

d/dx[1/(1 + x²)] = -2x/(1 + x²)²

Then

$\left(\dfrac1{1+x^2}y\right)' = 0$

$\dfrac1{1+x^2}y = C$

$y = C(1 + x^2)$

When x = 0, y = 3, so

$3 = C(1+0^2) \implies C=3$

$\implies \boxed{y = 3(1 + x^2) = 3 + 3x^2}$

7 0

2 years ago

Solve : <br><br>1) 5x – 15x²<br>2) 3x-9

8090 [49]

Answer:

1) 5x(1 - 3x)

2) 3(x - 3)

Step-by-step explanation:

<u>Question # 1:</u>

= 5x - 15x²

Take 5x common

= 5x(1 - 3x)

<u>Question # 2:</u>

= 3x - 9

Take 3 common

= 3(x - 3)

$\rule[225]{225}{2}$

Hope this helped!

4 0

2 years ago

Read 2 more answers

Is v23 rational or irrational

Alekssandra [29.7K]

Answer:

Assuming you mean the square root of 23, it is irrational.

6 0

4 years ago

Find the average rate of change for the given function from x=-1 to x=4

loris [4]

I believe the slope is 0 because it they are both horizontal lines? I may be wrong though.

6 0

4 years ago