Algebra - ACT Math

x2 + 5x – 6 = 0 factors to (x+6)(x**–**1) = 0.

Therefore, the two real solutions are x = **–**6 and x = 1. Their product is simply **–**6.

Use FOIL: (2g – 3h)(2g – 3h) = 4g2 – 6gh – 6gh + 9h2 = 4g2 – 12gh + 9h2

If we solve the first equation for 2x we find that 2x = 9 – y. If we solve the second equation for z we find z = –4 + y. Adding these two manipulated equations together we see (2x) + (y) = (9 – y)+(–4 + y).

The y’s cancel leaving us with an answer of 5.

We must find a common denominator and here they changed the first fraction by removing a negative from the numerator and denominator, leaving –11/(7 – x). We add the numerators and keep the same denominator to find the answer.

Factor the polynomial by choosing values that when FOIL'ed will add to equal the middle coefficient, 3, and multiply to equal the constant, 1.

x_2 – 6_x + 5 = (x – 1)(x – 5)

Because only (x – 5) is one of the choices listed, we choose it.

There are two ways to do this problem. The first way just involves plugging in –3 for x and solving 6〖(–3)〗2 + 16(–3) – 6, which equals 54 – 48 – 6 = 0. The second way involves factoring the polynomial to (6x – 2)(x + 3) and then plugging in –3 for x. The second way quickly shows that the answer is 0 due to multiplying by (–3 + 3).

Begin by multiplying both sides by (x – 1):

x2 – x = x – 1

Solve as a quadratic equation: x2 – 2x + 1 = 0

Factor the left: (x – 1)(x – 1) = 0

Therefore, x = 1.

However, notice that in the original equation, a value of 1 for x would place a 0 in the denominator. Therefore, there is no solution.

Generally, quadratic equations have two answers.

First, the equations must be put in standard form: 3x2 + 10x + 3 = 0

Second, try to factor the quadratic; however, if that is not possible use the quadratic formula.

Third, check the answer by plugging the answers back into the original equation.

If n=4, then 64(4/12)=64(1/3)=4. Then, 4=mŸ4(1+m)/(m+4). If 2 is substituted for m, then 4=2Ÿ4(1+2)/(2+4)=2Ÿ41/2=2Ÿ√4=2Ÿ2=4.

Since we have –2x and +2x in the equations, it makes sense to add the equations together to give 8y = 16 yielding y = 2. Then we substitute y = 2 into one of the original equations to get x = –2. So the solution to the system of equations is (–2, 2)

The y-intercept is the constant at the end of the equation. Thus for our equation the y-intercept is 7

To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

4_x_ + 3 = 5_x_ + 44; 3 = x + 44; –41 = x

To find the y-coordinate, substitute into one of the equations. Let's use _y_1:

y = 4 * –41 + 3 = –164 + 3 = –161

The center of our circle is therefore: (–41, –161).

Now, recall that the general form for a circle with center at (_x_0, _y_0) is:

(x - _x_0)2 + (y - _y_0)2 = _r_2

For our data, this means that our equation is:

(x + 41)2 + (y + 161)2 = 122 or (x + 41)2 + (y + 161)2 = 144

To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

To find the y-coordinate, substitute into one of the equations. Let's use :

The center of our circle is therefore .

Now, recall that the general form for a circle with center at is

For our data, this means that our equation is:

A line is an asymptote in a graph if the graph of the function nears the line as X or Y gets larger in absolute value.

The denominator cannot be zero, otherwise the function is indefinite. Therefore x cannot be –2 or –3.

The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by

The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.

The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.

The equation of the line is y – 4 = (3/4)(x – (–3))

Rearranging gives us: 3_x_ – 4_y_ = -25

Rewrite the equation of the circle in standard form to find its center:

Complete the square:

The center is .

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints and will have slope

,

so the tangent line has the opposite of the reciprocal of this, or , as its slope.

The tangent line therefore has equation

The graph of the equation is a circle with center .

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints and will have slope

,

so the tangent line has the opposite of the reciprocal of this, or , as its slope.

The tangent line therefore has equation