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If by + 40 you mean 40%, then the exponential growth rate would be:

1.4^x, where x = number of years.

It is 0.16 or 4/25 yay

Height = Sin(20) x 9 = 3.078

Area = Base x Height x 1/2

Area = 14 x 3.078 x 1/2

Area =21.5 cm^2

This is something called a PERFECT SQUARE TRINOMIAL (PST). It is, well, the square of a binomial. These are the two forms:

\((x+y)^2=x^2+2xy+y^2\)

\((x-y)^2=x^2-2xy+y^2\)

So, the factored form of your trinomial is \((x-y)^2\) 

You made an error in multiplying   (x - 5)(x + 2)   and   (x + 10)(2x + 1) .

(x - 5)(x + 2)   =   (x)(x) + (x)(2) + (-5)(x) + (-5)(2)   =   x² - 3x - 10

and

(x + 10)(2x + 1)   =   (x)(2x) + (x)(1) + (10)(2x) + (10)(1)   =   2x² + 21x + 10

Otherwise your method is correct. 

\(\qquad\ \frac{(x-5)(x+2)+(x+10)(2x-1)}{(x-2)(x+10)(x+2)} \\ =\\ \qquad\frac{(x^2-3x-10)+(2x^2+21x+10)}{(x-2)(x+10)(x+2)} \\ =\\ \qquad\frac{3x^2+18x}{(x-2)(x+10)(x+2)} \\ =\\ \qquad\frac{3x(x+6)}{(x-2)(x+10)(x+2)}\)

The LCD is actually just 3*4. This is because 12/6 = 2. When you multiply this to each fraction, you get the answer that the text gave.

This is what I did:

1. Factor out \({1\over x-2}\): \({1\over x-2}({x-5\over x+10}+{2x+1\over x+2})\)

2. Apply the cross-horizontal method: \({1\over x-2}({(x+2)(x+5)+(x+10)(2x+1)\over(x+10)(x+2)})\)

3. Simplify the numerator: \({1\over x-2}({3x^2+18x\over(x+10)(x+2)})\)

4. Factor out 3x from the numerator: \({1\over x-2}({3x(x+6)\over(x+10)(x+2)})\)

5. Multiply: \({3x(x+6)\over(x-2)(x+10)(x+2)}\)

Your error is in the second line wherein you applied FOIL. \(-5*2=-10\). You forgot to consider the signs 

a) If both f and g are even, then \(h(x)=f(g(x))=f(g(-x))=h(-x)\). EVEN

b) If both f and g are odd, then \(h(-x)=f(g(-x))=f(-g(x))=-f(g(x))=-(h(x))\). ODD

c) If f is even and g is odd, then \(h(-x)=f(g(-x))=f(-g(x))=f(g(x))=h(x)\). EVEN

d) If f is odd and g is even, then \(h(x)=f(g(x))=f(g(-x))=h(-x)\). EVEN

Thank you I am cured!!!

So, \(\triangle RST\) has two given interior angles, 60^o and 90^o. Since the sum of the interior angles of a triangle is 180^o, we can deduce that the remaining angle has a measure of 30^o. This is a special triangle in which the hypotenuse is double the shorter leg, and the longer leg is \(\sqrt{3}\) times the shorter leg. Since \(\overline{RS}\) is the longer leg and has a measure of  \(2\sqrt3\), the measure of \(\overline{ST}\) is \({2\sqrt3\over\sqrt3}=2\). This means that the measure of hypotenuse \(\overline{RT}\) is equal to \(2*2=4\). Now, \(\triangle QRT\) also has two given angles, 45^o and 90^o. By the same logic as before, the remaining angle has a measure of 45^o. This is also a special triangle, in which the two legs are equal, and the hypotenuse is \(\sqrt2\) times the leg. Since \(\overline{RT}\) and \(\overline{ RQ}\) are both legs of the triangle, their measures must be equal. Therefore, \(x=4\).

Since the center, focus, and vertex all lie on the line \(y=3\), the hyperbola opens left/right. This means that we use this form:

\({(x-h)^2\over a^2}-{(y-k)^2\over b^2}=1\)

Now, we substitute the values as follows:

     1. h is the x-coordinate of the center

     2. k is the y-coordinate of the center

     3. a is the distance from the center to the vertex

     4. b² = c² - a², wherein c is the distance from the center to the focus

Which means:

     1. \(h=3\)

     2. \(k=3\)

     3. \(a=\sqrt{(3-6)^2+(3-3)^3}=3\)

     4. \(c=\sqrt{(3-8)^2+(3-3)^2}=5\), \(b^2=25-9=16\)

Substituting, we get \({(x-3)^2\over9}-{(y-3)^2\over16}=1\), and that is your equation.

a  ÷  b  ÷  c   =   (a  ÷  b)  ÷  c

So...

\(\begin{array}\ \frac{8y^2-2y-3}{y^2-1}\div\frac{2y^2-3y-2}{2y-2}\div\frac{3-4y}{y+1}\,&=&\,\Big(\frac{8y^2-2y-3}{y^2-1}\div\frac{2y^2-3y-2}{2y-2}\Big)\div\frac{3-4y}{y+1}\\~\\ &=&\,\Big(\frac{8y^2-2y-3}{y^2-1}\cdot\frac{2y-2}{2y^2-3y-2}\Big)\div\frac{3-4y}{y+1}\\~\\ &=&\,\Big(\frac{8y^2-2y-3}{y^2-1}\cdot\frac{2y-2}{2y^2-3y-2}\Big)\cdot\frac{y+1}{3-4y}\\~\\ &=&\,\frac{8y^2-2y-3}{y^2-1}\cdot\frac{2y-2}{2y^2-3y-2}\cdot\frac{y+1}{3-4y}\\~\\ &=&\,\frac{(8y^2-2y-3)(2y-2)(y+1)}{(y^2-1)(2y^2-3y-2)(3-4y)} \end{array}\)

Then we can factor the numerator and denominators and cancel the common factors.

\(\begin{array}\ \frac{8y^2-2y-3}{y^2-1}\div\frac{2y^2-3y-2}{2y-2}\div\frac{3-4y}{y+1}\,&=&\,\frac{(8y^2-2y-3)(2y-2)(y+1)}{(y^2-1)(2y^2-3y-2)(3-4y)}\\~\\ &=&\,\frac{(2y+1)(4y-3)(2)(y-1)(y+1)}{(y+1)(y-1)(y-2)(2y+1)(-1)(4y-3)}\\~\\ &=&\,\frac{(2)}{(y-2)(-1)} \\~\\ &=&\,-\frac{2}{y-2}\qquad\text{and }\,y\neq-1,1,-\frac12,\frac34 \end{array}\)

So in this case, flipping the latter two fractions and changing the signs to multiplication will work. But note that if problem were given as  a ÷ (b ÷ c) , only the middle fraction would end up being flipped. So it is easiest to do them one at a time. Just remember that if there are no parenthesees given, it means the same as  (a ÷ b) ÷ c  . 

give the initial value and put the growth factor as a percent?

tysm

FV = PV [1 + R]^N

FV = 20,000 x [1 - 0.12]^3

FV = 20,000 x    0.88^3

FV = 20,000 x     0.681472

FV = $13,629.44 - the value of the car after 3 years.

happy (late) birthday!

Tan(x) = Opposite side / Adjacent side

Not quite:

Factor the following:
2 y^2 - 5 y - 3

Factor the quadratic 2 y^2 - 5 y - 3. The coefficient of y^2 is 2 and the constant term is -3. The product of 2 and -3 is -6. The factors of -6 which sum to -5 are 1 and -6. So 2 y^2 - 5 y - 3 = 2 y^2 - 6 y + y - 3 = y (2 y + 1) - 3 (2 y + 1):
y (2 y + 1) - 3 (2 y + 1)

Factor 2 y + 1 from y (2 y + 1) - 3 (2 y + 1):

=(2y + 1) (y - 3)

h=2f, for f       divide both sides by 2

f = h / 2

I don't have enough tiem to do it properly but here are some ideas.

The area of sector is 60/360 * area of the whole circle.

the quad is a rhombus which means both are equilateral triangles.

Find the area of one triangle. Might want to find height first though not strictly necessary.

subtract and then double to get shaded area.

Sorry do not have time to write instructions well.

Thanks so much CPhill!

Are you sure all the bases are correct?

x = 8 / [8 + 12] x 24 = 9.6

Did you find how to this this? I also need help on this question....

IT IS AND ALWAYS WILL BE 180°