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3 years ago

9/10(x+2/3)=4 1/2 What is x in this equation

Mathematics

1 answer:

dexar [7]3 years ago

6 0

Answer:

x = 13/3

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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2 years ago

Kofi wants to find the most-liked snacks to put in the vending machine at his office so the most people will buy from it.

Whitepunk [10]

Answer:

The correct option is B

Step-by-step explanation:

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3 years ago

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One 16-ounce bottle of an energy drink has an average of 500 mg of caffeine with a standard deviation of 25 mg. In a carton cont

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<span>In this case, every bottle has 500 mg of caffeine with a standard deviation of 25 mg. The equation of a bottle will be: 500 +/- 25mg caffeine per bottle.

To calculate a carton containing 30 bottles, you just need the equation with 30. It will become: (500x30) +/- (25x30)mg= 15000 +/- 750 mg of caffeine per carton</span>

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3 years ago

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Sliva [168]

Step-by-step explanation:

i don't understand the ques

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2 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

2 years ago