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We know

d   =   r * t              Divide both sides of this equation by  r  to get

d / r   =   t

So....

t  =  d / r

Factor the following:
18 x^3 - 8 x y^2

Factor common terms out of 18 x^3 - 8 x y^2.
Factor 2 x out of 18 x^3 - 8 x y^2:
2 x (9 x^2 - 4 y^2)

Express 9 x^2 - 4 y^2 as a difference of squares.
9 x^2 - 4 y^2 = (3 x)^2 - (2 y)^2:
2 x (3 x)^2 - (2 y)^2
Factor the difference of two squares. (3 x)^2 - (2 y)^2 = (3 x - 2 y) (3 x + 2 y):

2x (3x - 2y) (3x + 2y)

I'm no expert in Statistics, but I think the z-score is:

z =[ X - M] / S, where X = x-score, M = Mean, S = Standard Deviation. If this is right, then:

z-score =[130 - 100] / 15 = 2

You have to find cos(x) from sin(x) =4/5, which would be:cos(x) =3/5. Then;

You have to find cos(y) from sin(y) =5/13, which would be:cos(y) =12/13. Then:

You have to use this formula: sin(x + y) = sin x cos y + cos x sin y=4/5*12/13 + 3/5*5/13 =63/65 

After you post the question, then try to "edit" it and use the "scissors" in the top left-hand corner to "cut" it. I think it might work.

How can you delete a question? I have never been able to do it.

I good thanks

Good morning to you, but good night to me.

Here is Hong Kong, and it is 22:28PM. 

My day has been good, favourite lessons all in a day... :D

Programming, Chemistry, Physics.... :D

For 6), never heard of it! Definitely I have made a right choice making an account here 1 or 2 years ago! :D

For each functon find the following (if they exist).

     a. End Behavior including the equations of horizontal or slant asymptotes.

     b. Vertical Asymptote(s). Distinguish between VA and Holes.

$y=\dfrac{-3x^3}{x^3-4x}\\ y=\dfrac{-3x^3}{x(x-2)(x+2)}\\~\\ so\;\;x\ne\pm2,\;\;\;x\ne0\\~\\ y=\dfrac{-3x^3}{x^3-4x}\\ $

$y=\dfrac{-3x^3\div x^3}{(x^3-4x)\div x^3}\\ y=\dfrac{-3}{1-\frac{4}{x^2}}\\ \displaystyle\lim_{x\rightarrow 0^\pm}y=\frac{-3}{1-\infty}=0 \quad \text{Hole at (0,0)} \\\displaystyle\lim_{x\rightarrow \pm\infty}y=\frac{-3}{1-0}=-3\\~\\ \text{Let }\delta\;\;\text{ be a miniscule positive number.}\\ \displaystyle\lim_{x\rightarrow2^+}y=\frac{-3}{1-(1-\delta)}=\frac{-3}{\delta}=-\infty\\ \displaystyle\lim_{x\rightarrow-2^-}y=\frac{-3}{1-(1-\delta)}=\frac{-3}{\delta}=-\infty\\ \displaystyle\lim_{x\rightarrow2^-}y=\frac{-3}{1-(1+\delta)}=\frac{-3}{-\delta}=\infty\\ \displaystyle\lim_{x\rightarrow-2^+}y=\frac{-3}{1-(1+\delta)}=\frac{-3}{-\delta}=\infty\\ $
 

Hole at (0,0)

Vertical asymptotes at  x=+2 and x=-2

Horizontal asymptotes at  y=-3

Here is the graph

A triangle has side lengths measuring 30, 40 and 50 units. What is the length of its shortest altitude, in units?

Here is the picture.

The altitudes are 30, 40, and h

h is the smallest because h is not the hypotenuse of either of the 2 created triangles.

$x^2+h^2=30^2 \qquad \qquad h^2+(50-x)^2=40^2\\ x^2+h^2=900 \qquad \qquad h^2+2500-100x+x^2=1600\\ x^2+h^2=900 \qquad \qquad h^2+x^2=1600-2500+100x\\ x^2+h^2=900 \qquad \qquad x^2+h^2=100x-900\\ so\\ 100x-900=900\\100x=1800\\x=18\\~\\ h^2=900-x^2\\h^2=900-18^2\\h^2=576\\h=\sqrt{576}\\h=24\\$

So the smallest altitude is 24

cos(ABM) = ?

$AM=MC=BM=r(\text{circle })$

$\tan{ACB}=\frac{10}{24}$

$\frac{\sin(ACB)}{r}=\frac{\sin(90^{\circ}-ABM)}{r}\\ \sin(ACB)=\cos(ABM)\\ \cos(ABM)=\sin(\arctan(\frac{10}{24}))=0.384615384615...\\ ABM=\arccos(\frac{5}{13})=67.380135051983^{\circ }$

 heureka

thank you :)

I'm going to assume that there is an original equation of y=(-1/2)x+b with a y-value of 2 and an x-value of -3, solve for b.

$2=-\frac{1}{2}(-3)+b$

Multiply.

$2=\frac{3}{2}+b$

Subtract b from both sides.

$\frac{4}{2}-\frac{3}{2}=b$

$b=\frac{1}{2}$

Now plug back into original equation.

$y=-\frac{1}{2}x+\frac{1}{2}$

In a problem with mx+b format, m is the slope, or in this problem $m=-\frac{1}{2}$.

Answer it all if possible, you can do it in parts, like A [send] B [send] and so on...

For this, plug values 2 and 4 in for x.

$f(2)=(4)^2-1, f(4)=(4)^4-1$

$f(2)=16-1, f(4)=256-1$

$f(2)=15, f(4)=255$

For the average, add the two values and divide by 2.

$\frac{15+255}{2}=\frac{270}{2}=135$

So the average increase is 135 students.

To find that you have to complete the square. The formula for that is $(\frac{b}{2})^2$.

In this equation b is -1.

$(-\frac{1}{2})^2=\frac{1}{4}$

So it would be $x^2-x+\frac{1}{4}$.

S = F/[1 -R]

S = 2 / [1 - 1/18]

S = 2 / [17/18]

S = 2 x [18/17]

S = 36 / 17

S = 2 2/17

f(x)=10*5.5^x

f(1)=10*5.5 = 55

f(2)= 10*  30.25 = 302.5

average rate of change = rise/run = (302.5-55)/(2-1) = 247.5

Thank you!

Fishing reel =r

Fishing Rod=R

3r + 5R =488, 

 6r + 6R =588, solve for r, R

Solve the following system:
{3 r + 5 R = 488 | (equation 1)
6 r + 6 R = 588 | (equation 2)

Swap equation 1 with equation 2:
{6 r + 6 R = 588 | (equation 1)
3 r + 5 R = 488 | (equation 2)

Subtract 1/2 × (equation 1) from equation 2:
{6 r + 6 R = 588 | (equation 1)
0 r+2 R = 194 | (equation 2)

Divide equation 1 by 6:
{r + R = 98 | (equation 1)
0 r+2 R = 194 | (equation 2)

Divide equation 2 by 2:
{r + R = 98 | (equation 1)
0 r+R = 97 | (equation 2)

Subtract equation 2 from equation 1:
{r+0 R = 1 | (equation 1)
0 r+R = 97 | (equation 2)

Collect results:
r = $1.00          and         R=$97.00
 

iDK ASK A MODERATOR OR SOMETHING

This isn't really a trick right? It is just some subtraction and addition.......

It's like those pick a random number......now subtract you original number.

It just cancels out...

ax² + bx + c = 0

To solve this problem, we need to use the formula X₁*X₂ = c ÷ a

(√3-1) * X₂ = 4 ÷ 2

 X₂ = 2/ √3-1

Using that, we can figure out k, 

Using this 

X₁ + X₂ = -a/b

(2/√3-1) + (√3-1) = -(-2k/2)

We got k = 2√3

Can someone check over my work? Not sure about the answer, pretty sure the solution is right...

Find the derivative of the following via implicit differentiation:
d/dx(y) = d/dx((x^(1/3) + sqrt(x) + x)/x^2)

The derivative of y is y'(x):
y'(x) = d/dx((x^(1/3) + sqrt(x) + x)/x^2)

Use the product rule, d/dx(u v) = v ( du)/( dx) + u ( dv)/( dx), where u = 1/x^2 and v = x + sqrt(x) + x^(1/3):
y'(x) = (x^(1/3) + sqrt(x) + x) d/dx(1/x^2) + (d/dx(x^(1/3) + sqrt(x) + x))/(x^2)

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = -2: d/dx(1/x^2) = d/dx(x^(-2)) = -2 x^(-3):
y'(x) = (d/dx(x^(1/3) + sqrt(x) + x))/x^2 + (x^(1/3) + sqrt(x) + x) (-2)/x^3

Differentiate the sum term by term:
y'(x) = -(2 (x^(1/3) + sqrt(x) + x))/x^3 + (d/dx(x^(1/3)) + d/dx(sqrt(x)) + d/dx(x))/x^2

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 1/3: d/dx(x^(1/3)) = d/dx(x^(1/3)) = x^(-2/3)/3:
y'(x) = -(2 (x^(1/3) + sqrt(x) + x))/x^3 + (d/dx(sqrt(x)) + d/dx(x) + 1/(3 x^(2/3)))/x^2

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 1/2: d/dx(sqrt(x)) = d/dx(x^(1/2)) = x^(-1/2)/2:
y'(x) = -(2 (x^(1/3) + sqrt(x) + x))/x^3 + (1/(3 x^(2/3)) + d/dx(x) + 1/(2 sqrt(x)))/x^2

The derivative of x is 1:
y'(x) = -(2 (x^(1/3) + sqrt(x) + x))/x^3 + (1/(3 x^(2/3)) + 1/(2 sqrt(x)) + 1)/x^2

Simplify the expression:
y'(x) = (1 + 1/(3 x^(2/3)) + 1/(2 sqrt(x)))/x^2 - (2 (x^(1/3) + sqrt(x) + x))/x^3 - Courtesy of "Mathmatica 11 Home Edition"