All categories

2 years ago

If y is directly proportional with x and y=24 when x=6. What is the value of y when x=48

Mathematics

1 answer:

Thepotemich [5.8K]2 years ago

7 0

Answer:

192

Step-by-step explanation:

Multiply 48 by 24, then divide by 6

You might be interested in

Pat said that the amount of money he has in his pocket is greater than or equal to $5.75 but less than $13.00. Which number line

blsea [12.9K]

Draw a line that starts at 5.75 and ends at 12.99

7 0

2 years ago

A fair coin is tossed three times in succession. Find the probability if getting exactly one tail

frosja888 [35]

Answer:

If a fair coin is tossed three times, sample = 2^3 = 8 (hhh, hht, hth, thh, htt, tht, tth, ttt) where h=head and t=tail.

Probability of getting one:

Outcomes with one head (h) = 3

P(1 h) = 3/8 = 0.375

3 0

3 years ago

What is the answer to this math equation? <br><br> -3y < -14

Illusion [34]

Answer:

y = 5

Step-by-step explanation:

-3 * 5 = -15

-15 < -14

5 0

2 years ago

Find the product 8/6n - 4(9n^2 -4)

Brrunno [24]

Answer:

- (27n^3 - 12n - 1) over 3n

Step-by-step explanation:

4 0

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

2 years ago