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3 years ago

Rewrite the proportion 7:21 = 3:9 as a proportion using fractions.

Mathematics

2 answers:

Inessa [10]3 years ago

4 0

Answer:

1/3 ig

Step-by-step explanation:

VARVARA [1.3K]3 years ago

4 0

Answer: In this problem you want to rewrite the proportion 7:21 to 3:9.

First, a proportion is a relation between two quantities, this is if the proportion of A to B is 7:21, then you can represent it in a equation as:

7*A = 21*B

If you divide by the same number in both sides of the equation, you dont affect the equality, then if i divide both sides by 7, the equation turns into:

7*A/7 = 21*B/7

A = (21/7)*B = 3*B

Now if i multiply both sides by 3.

3*A = 3*(3*B) = 9*B

3A = 9*B

so now the proportion is 3:9.

Also, if you want to see if two proportions are equivalent, you need to see that quotients are the equal.

In this case 21/7 = 3, and 9/3 = 3.

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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!

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svlad2 [7]

Answer:

Step-by-step explanation:

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Compare 2.3 and 2.03

Andru [333]

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Logan and his brother plan to make spherical glass ornaments to decorate their living room. Logan decides to completely fill the

mixer [17]

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