Mathhemathh

oh no did i do something wrong

EDIT: Yes I did. Zero is not positive xD

So, the only way for the solutions to be rational is if the discriminant ($\sqrt{b^2-4ac}$) is rational. Substituting a and c, we get $\sqrt{121-20c}$. Since this has to be rational, $121-20c$ has to be a perfect square. The only possible values of c wherein this is a perfect square are 0 (edit: 6) and 2, resulting in 1 and 81 respectively. Getting the product of 6 and 2, we get, well, 12.

CPhill's cognitive power is unmatched.

For the first question:

When an equation has infinitely many solutions, it usually can be reduced to 0 = 0. So, what I usually do is try to get as close as possible, like this:

$2(x+b)=ax+c$

$2x+2b=ax+c$

Well... this was as close as I got. After this, think, what value does c have to be for this equation to become 0 = 0? If you think about it, the two expressions are already very similar. a could be equal to 2, and c could be equal to 2b. To test this, substitute c for 2b to see if it leads to any contradictions:

$2x+2b=ax+2b$

$2x=ax$

$2=a$

This is what we assumed before, that a = 2. Therefore, c = 2b, the answer is d.

For the second question:

If you look at the expression $x^2-4xy+4y^2$, it is a perfect square trinomial, equal to $(x-2y)^2$. Also, if you simplify $3x-6y=9z$, it becomes $x-2y=3z$. Substituting that into our PST, we get $(3z)^2=9z^2$. The answer is c.

First, we illustrate the problem:

I - the incenter (center of inscribed circle)

Then, draw lines like this:

Notice that the triangle has been split into 3 smaller ones, each with a height of the radius, and with a base of the sides of the large triangle. Which means that we can say:

$a\triangle ABC=a\triangle AIB+a\triangle BIC+a\triangle CIA$

$a\triangle ABC={5r\over 2}+{6r\over 2}+{7r\over 2}$

$a\triangle ABC =9r$

Now, we also know that $a\triangle ABC=\sqrt{s(s-AB)(s-AC)(s-BC)}$ wherein $s={AB+AC+BC\over2}$. This is called Heron's Formula. Substituting the values, we get:

$\sqrt{9*4*3*2}=9r$

$6\sqrt{6}=9r$

$r={2\sqrt{6}\over3}$

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 $(x-y)^2$ 

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

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.

So...

We are trying to find the maximum value of the function $Y(p)={900p\over2+p}$ within the domain $x\ge0$. This is the graph of a hyperbola, and it is positive. So, in order to find the maximum value of the function, we have to get its horizontal asymptote. An asymptote is a line that a graph approaches but never touches.  The rules are as follows:

1. If the numerator's degree is less than the denominator's degree, then the horizontal asymptote is y = 0.

2. If the numerator's degree is equal to the denominator's degree, the the horizontal asymptote is the quotient of the coefficients of the highest degree terms.

3. If the numerator's degree is greater than the denominator's degree, then it does not have a horizontal asymptote.

The degree of 900p is 1, and the degree of 2+p is 1. Therefore, the horizontal asymptote is $y={900\over1}$, or just $y=900$. This means that the maximum value of the function is 900.

What this actually means is that the function approaches 900 as you go farther along the x-axis, or, as you increase the amount of pesticide. So, theoretically, if you had infinite pesticide, you would have 900kg of extra yield. But... infinity is tricky. Just trust me, the answer is 900.

Do we have to specifically substitute values to find the answer, or can we use a different method?

Observe that the number of disaster-caused deaths prevented is doubled from the first statement to the second statement (25% > 50%) and the number of people saved per year goes from 0.25% to 1.9%. This means that doubling the deaths prevented results in the number of people saved multiplied by a factor of ${0.019\over0.0025}$, or 7.6. As an equation, this is $2d=7.6D$, wherein d is the number of disaster-caused deaths prevented, and D is the number of total deaths prevented.

Now, observe that the number of disaster-caused deaths prevented is tripled from the first statement to the third statement. To find how much the number of total deaths prevented, we just need to manipulate the equation we derived earlier:

${3\over2}*2d={3\over2}*7.6D$

$3d=11.4D$

By tripling the number of disaster-caused deaths prevented, we multiply the number of total deaths prevented per year by 11.4. Now, since d tripled from the first statement to the third,  we must multiply 0.25% by 11.4. Doing this, we get 2.85% of deaths prevented each year.

(tbh i'm not very confident with this answer, but it was the best I could do)

1. Distributing the 4 and the -7, we get $4x^2-8x+8-7x^3+21x-7$. Combining like terms, we get $-7x^3+4x^2+13x+1$. The coefficients are -7, 4, and 13. The sum of their squares is 49+16+169 = 234.

2. Multiplying the two polynomials (multiply each possible pair of terms and add), we get $2x^4+ax^3+7x^2-6x^3-3ax^2-21x+8x^2+4ax+28=2x^4+(a-6)x^3+(15-3a)x^2+(4a-21)x+28$Since the coefficients of the two expressions (this expression and the second expression you gave) have to be equal for the polynomials to be equal, $a-6=-11$ and $15-3a=30$ and $4a-21=-41$. Solving for a, we get a = -5.

Marginal Profit = Marginal Revenue - Marginal Cost

Marginal Cost = C'(x) = 0.02

Revenue = No. of Items sold * Price = x * (3 -  (x/1000)) = $-{x^2\over1000}+3x$

Marginal Revenue = R'(x) = $-{x\over500}+3$

Marginal Profit = $-{x\over500}+3-0.02$ = $2.98-{x\over500}$

The answer is letter B