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2 years ago

9

Brandon and his friend combined their allowance of $900 each, so they could buy a movie a 2-for-1 movie ticket for $1200. They b

ought $200 bags of popcorn with the remaining money and split the amount evenly between them. How many containers of popcorn did each boy get?

1 answer:

inysia [295]2 years ago

5 0

Answer:

1.5 bags

Step-by-step explanation:

They each had 900 dollars, so when combined they have 1800 dollars. They bought a movie ticket for 1200 dollars, so they now have 1800-1200=600 dollars.

600/2 is 300. Each bag of popcorn costs 200 dollars. So they could each buy 1.5 bags of popcorn.

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In how many ways can a cricket team of eleven people be chosen out of a batch of 15 players?

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2 years ago

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

mote1985 [20]

Answer:

$\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

Step-by-step explanation:

To find the derivative of the function $y(x)=\ln \left(\frac{x}{x^2+1}\right)$ you must:

Step 1. Rewrite the logarithm:

$\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 2. The derivative of a sum is the sum of derivatives:

$\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }}={\left(\left(\ln{\left(x \right)}\right)^{\prime } - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }\right)$

Step 3. The derivative of natural logarithm is $\left(\ln{\left(x \right)}\right)^{\prime }=\frac{1}{x}$

${\left(\ln{\left(x \right)}\right)^{\prime }} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }={\frac{1}{x}} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 4. The function $\ln{\left(x^{2} + 1 \right)}$ is the composition $f\left(g\left(x\right)\right)$ of two functions $f\left(u\right)=\ln{\left(u \right)}$ and $u=g\left(x\right)=x^{2} + 1$

Step 5. Apply the chain rule $\left(f\left(g\left(x\right)\right)\right)^{\prime }=\frac{d}{du}\left(f\left(u\right)\right) \cdot \left(g\left(x\right)\right)^{\prime }$

$-{\left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }} + \frac{1}{x}=- {\frac{d}{du}\left(\ln{\left(u \right)}\right) \frac{d}{dx}\left(x^{2} + 1\right)} + \frac{1}{x}\\\\- {\frac{d}{du}\left(\ln{\left(u \right)}\right)} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- {\frac{1}{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}$

Return to the old variable:

$- \frac{1}{{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- \frac{\frac{d}{dx}\left(x^{2} + 1\right)}{{\left(x^{2} + 1\right)}} + \frac{1}{x}$

The derivative of a sum is the sum of derivatives:

$- \frac{{\frac{d}{dx}\left(x^{2} + 1\right)}}{x^{2} + 1} + \frac{1}{x}=- \frac{{\left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right)}}{x^{2} + 1} + \frac{1}{x}=\frac{1}{x^{3} + x} \left(x^{2} - x \left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right) + 1\right)$

Step 6. Apply the power rule $\frac{d}{dx}\left(x^{n}\right)=n\cdot x^{-1+n}$

$\frac{1}{x^{3} + x} \left(x^{2} - x \left({\frac{d}{dx}\left(x^{2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(x^{2} - x \left({\left(2 x^{-1 + 2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x \frac{d}{dx}\left(1\right) + 1\right)\\$

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Thus, $\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

3 0

3 years ago

olivia usues 1/3 cup of orange juice for every 2/3 cup of pineapple juice to make a fruit drink. find the number of cups olivia

4vir4ik [10]

I’m sort of confused. Do you mean how many cups of fruit juice she’ll get if she uses 2 cups of pineapple juice? If so it should be 3.

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3 years ago

What is the quadratic equation if the solutions are -9+ √65 /2, -9- √65 /2

liberstina [14]

Answer:

$4x^2+72x+259=0$

Step-by-step explanation:

If $x_1$ and $x_2$ are the solutions to the quadratic equation, then this equation can be written as

$(x-x_1)(x-x_2)=0$

In your case,

$x_1=-9+\dfrac{\sqrt{65}}{2}\\ \\x_1=-9-\dfrac{\sqrt{65}}{2}$

Then the equation is

$\left(x-\left(-9+\dfrac{\sqrt{65}}{2}\right)\right) \left(x-\left(-9-\dfrac{\sqrt{65}}{2}\right)\right)=0\\ \\\left(x+9-\dfrac{\sqrt{65}}{2}\right)\left(x+9+\dfrac{\sqrt{65}}{2}\right)=0\\ \\x^2+9x+\dfrac{\sqrt{65}}{2}x+9x+81+\dfrac{9\sqrt{65}}{2}-\dfrac{\sqrt{65}}{2}x-\dfrac{9\sqrt{65}}{2}-\dfrac{65}{4}=0\\ \\x^2+18x+\dfrac{259}{4}=0\\ \\4x^2+72x+259=0$

4 0

3 years ago