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4 years ago

Find the original price of a pair of shoes if the sale price is $ 11 after a 75 % discount.

Mathematics

2 answers:

tekilochka [14]4 years ago

7 0

The answer is either 33 or 44.

Anna11 [10]4 years ago

3 0

The ORIGINAL price is $44. You are saving $33.

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The product of 2 and y is at least 10. solving inequalities

White raven [17]

2•y>10
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3 years ago

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Dustin and Melanie are playing a game, where two standard, six-sided number cubes are rolled, and the sum of their outcome is fo

Iteru [2.4K]

Answer:

Melanie.

Step-by-step explanation:

The possible ways of getting a 6 are (3,3), (2,4), (4,2), (1,5) and (5,1).

That is 5 ways.

The possible ways of getting a 7 are (1,6), (6,1), (5,2),( 2,5), (3, 4), (4,3).

That is 6 ways.

So Melanie made the better decision.

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3 years ago

The number a point corresponds to on a number line is called it's

cestrela7 [59]

It's called a coordinate

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3 years ago

Great! You will purchase snacks for the painting class. You have a budget of $40. You want to buy fruit and granola bars.

goldfiish [28.3K]

Forgot the question hahaha

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3 years ago

Find dy/dx by implicit differentiation for ysin(y) = xcos(x)

tatyana61 [14]

Answer:

$\frac{dy}{dx}=\frac{\cos(x)-x\sin(x)}{\sin(y)+y\cos(y)}$

Step-by-step explanation:

So we have:

$y\sin(y)=x\cos(x)$

And we want to find dy/dx.

So, let's take the derivative of both sides with respect to x:

$\frac{d}{dx}[y\sin(y)]=\frac{d}{dx}[x\cos(x)]$

Let's do each side individually.

Left Side:

We have:

$\frac{d}{dx}[y\sin(y)]$

We can use the product rule:

$(uv)'=u'v+uv'$

So, our derivative is:

$=\frac{d}{dx}[y]\sin(y)+y\frac{d}{dx}[\sin(y)]$

We must implicitly differentiate for y. This gives us:

$=\frac{dy}{dx}\sin(y)+y\frac{d}{dx}[\sin(y)]$

For the sin(y), we need to use the chain rule:

$u(v(x))'=u'(v(x))\cdot v'(x)$

Our u(x) is sin(x) and our v(x) is y. So, u'(x) is cos(x) and v'(x) is dy/dx.

So, our derivative is:

$=\frac{dy}{dx}\sin(y)+y(\cos(y)\cdot\frac{dy}{dx}})$

Simplify:

$=\frac{dy}{dx}\sin(y)+y\cos(y)\cdot\frac{dy}{dx}}$

And we are done for the right.

Right Side:

We have:

$\frac{d}{dx}[x\cos(x)]$

This will be significantly easier since it's just x like normal.

Again, let's use the product rule:

$=\frac{d}{dx}[x]\cos(x)+x\frac{d}{dx}[\cos(x)]$

Differentiate:

$=\cos(x)-x\sin(x)$

So, our entire equation is:

$=\frac{dy}{dx}\sin(y)+y\cos(y)\cdot\frac{dy}{dx}}=\cos(x)-x\sin(x)$

To find our derivative, we need to solve for dy/dx. So, let's factor out a dy/dx from the left. This yields:

$\frac{dy}{dx}(\sin(y)+y\cos(y))=\cos(x)-x\sin(x)$

Finally, divide everything by the expression inside the parentheses to obtain our derivative:

$\frac{dy}{dx}=\frac{\cos(x)-x\sin(x)}{\sin(y)+y\cos(y)}$

And we're done!

5 0

3 years ago