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3 years ago

6

Tavon has a $50 gift card that’s very loses $2 for each 30-day period it is not used. He has a $40 card that’s loses $1.50 for e

ach 30-day period it is not used.

1 answer:

sasho [114]3 years ago

3 0

Answer:

Hello, thank you for your word problem. However, your word problem doesn't have a question to it. Please double check your answer. Thank you!

Step-by-step explanation:

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A group of seven musicians includes three violinists and four harpists. An instructor randomly selects one of the musicians to p

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They are dependent events. Event A is choosing one musician while Event B is choosing another musician. THey are ALL musicians, no matter what instrument they play, we just know that there are 7 musicians in total. Therefore, the events are dependent because after choosing one musician, you would have 6 musicians left, instead of 7.

Step-by-step explanation:

3 0

3 years ago

A factory that makes granola bars uses 1⁄6 of a barrel of raisins in each batch. Yesterday, the factory used 5⁄6 of a barrel of

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Answer:

5 Batches.

Step-by-step explanation:

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5 0

3 years ago

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$

We use the initial condition $y(0)=-12$ to find the value of C.

$-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}$

Therefore,

$y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}$

4 0

3 years ago