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3 years ago

MEDAL

2 answers:

5 0

The statement <span>The coefficient of x^k y^n-k in the expansion of (x+y)^n equals (n-k / k) is true. This will show the standard formula and the expansion of it. We all know that it can still be expanded based on the power or degree of the terms.</span>

balandron [24]3 years ago

4 0

Answer: the answer is FALSE. Just took this on a p e x

Step-by-step explanation:

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Create a problem that would have the answer 3x+12

oee [108]

Answer:

Problem:

Consider a rectangle such that the length of the rectangle is 12 more than thrice its width. Find a formula for the length in terms of its width. Take width as 'x'.

Step-by-step explanation:

Consider a rectangle such that the length of the rectangle is 12 more than thrice its width. Find a formula for the length in terms of its width.

Let the width be 'x'.

Therefore, as per question, length is 12 more than thrice the width.

Thrice the width means $3x$ . 12 more means adding 12 to the result.

Therefore, the length of the rectangle is $3x+12$

So, the above question expresses the length of the rectangle as $3x+12$ which is the required answer.

7 0

3 years ago

What is the measure for the following<<br> M

kumpel [21]

m= if i get the Diagram i will answer

3 0

3 years ago

A university has 34,890 students. In a random sample of 270 students, 18 speak three or more languages. Predict the number of st

Nady [450]

The answer to your specific question is $126$ students.

6 0

4 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

There is maybe be answer two

dolphi86 [110]

I believe it would be the first choice

Hope that helps!

4 0

2 years ago

Read 2 more answers