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(A,B) = (5,-4).

f(x)  =  5 / (x - 2)

g(x)  =  2x / (x² - 4)

f(x) + g(x)  =  5 / (x - 2)  +  2x / (x² - 4)     =  5 / (x - 2)  +  2x / [ (x - 2)(x + 2) ]

                                     =  [ 5(x + 2) ] / [ (x - 2)(x + 2) ]  +  2x / [ (x - 2)(x + 2) ]

                                     =  [ 5(x + 2) + 2x ] / [ (x - 2)(x + 2) ]

                                     =  [ 5x + 10 + 2x ] / [ (x - 2)(x + 2) ]

                                              =  [7x + 10] / [ (x - 2)(x + 2) ]

(f + g)(-1)  =  [ 7 · -1 + 10 ] [ (-1 -2)(-1 + 2) ]  =  3 / -3  =  -1

2(7x+24y)=x^2+y^2

x^2-14x=48y-y^2

x(x-14)=y(48-y) 

(x,y)=(0,0) , (14,48) , (0,48),(14,0) etc.. 

This is an interesting problem because the first equation:  3x² + 4y² + 3x - 2y + 3 =  0  does not have a real-valued graph.

If you attempt to place this graphing form:      3x² + 3x  + 4y² - 2y + 3 =  0 

                                                                        3(x² + x)  +  4(y² - ½y)  =   -3

Completing the squares:                                 3(x² + x + ¼)  +  4(y² - ½y + 1/16)  =  -3 + 3/4 + 1/4

                                                                        3(x + ½)² + 4(y -  ¼)²  =  -2

This is not a real-valued function.

So, the points of intersection do not exist in a real-valued plane.

Equate them.....because when they are equal is when they have an intersection/common point

3x^2+4y^2+3x−2y+3=3x^2−2y^2+3x−y+3      delete 3x^2   and 3x from both sides

4y^2 -2y+3 = -2y^2-y+3      gather like terms on one side

6y^2 -y = 0

y(6y-1) = 0      y = 0    or 1/6      now that you have values for 'y'   sub back into one of the original equations to fing the corresponding 'x' values.

SO there is 2 points of intersection.

x^2 -x   = x (x-1)    that's about it 

If the ratio is 4:3, there are 7 parts.

The x-distance from (-2,-5) to (5,-3)  is  5 - -2  =  7     --->     (1/7) x 7  =  1     --->     move 4 x 1 spaces to the right of -2, you end at 2

The y-distance from (-2,-5) to (5,-3)  is  -3 - -5  =  2    --->     (1/7) x 2  =  2/7  --->     move 4 x (2/7)  space up from -5, you end at -3 6/7

The point is (2, -3 6/7)

Let  n  =  number of terms

Let  t₁  =  the first term

Let  t_n  =  the n^th term

Let  d  =  the common difference

Let  S  =  the sum of the terms

Two formulas:  S  =  n · (t₁ + t_n) / 2                        t_n  =  t₁ + (n - 1) · d

t_n  =  t₁ + (n - 1) · d     --->      t_n  =  18 + (20 - 1) · 11     --->     227

S  =  n · (t₁ + t_n) / 2     --->     S  =  20 · (18 + 227) / 2     --->     2450

Since all the cards are dealt out, x must evenly divide into 54.

The numbers that are factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54.

So, when x = 1, y = 54; x = 2, y = 27; x = 3, y = 18; x = 6, y = 9; x = 9, y = 6; x = 18, y = 3; x = 27, y = 2; and x = 54, y = 1.

For x to be at least 2 and y to be at least 5, the possibilities are (2, 27), (3,18), (6,9), (9,6).

(7 *77   + 5 * x )/ (5+7)  = 82

x= 89     scores needed on next 5 exams

This is very simple.

All you have to know is  tanx = sinx / cosx

Nvm, I solved it

(e^(x^3/6))/(e^(x^2/2))

(e^(x^3   /6))   /   (e^(x^2    /2))

\(\frac{e^{\frac{x^3}{6}}} {e^{\frac{x^2}{2}}}\\~\\ =\frac{e^{\frac{x^3}{6}}} {e^{\frac{3x^2}{6}}}\\~\\ =e^{\frac{x^3-3x^2}{6}} \\~\\ \)

7/8 of the recipe is water     7/8 x 1gal = 7/8 gallon of water needed

4 quarts X 7/8  = 28/8 quarts of water  = 3 1/2 q

LCM of 4 6 and 10 is     60    so every 60th crab pot has 3 balloons    600/60 = 10

Slope = rise/run  = y2-y1   /   x2-x1

                            -7-2      / 13 - -2

                             -9    / 15    =  -3/5

x-y = 36

x = 2y +4

sub into the first equation

2y+4 - y = 36

y = 32

then    x = 2y+4 = 68  

IF I understand the question correctly you can use this formula.
Since the difference between them is 36, we let X be the large number and y the small. Therefore X-Y=36. Since X is 4+2*y we can substitute X in the equation so X-Y=36 Becomes =4+2*Y-Y=36. Solve by 2*y-y=36-4=>y=32. Since x is 4+2*y we have... X=32*2+4=> X=68

Thanks

If you mean how many non-foreign stamps he collected the answer would be 250*(1-0.76)=> 60

I would assume it is X=20/Y.

As an example if X=5 and Y=4 you have 5*4=20. Conversely 20/5=4 and 20/4=5.

A circle is inscribed in a right isosceles triangle whose legs are 8 inches long. How many square inches are in the area of the shaded portion? Express your answer as a decimal rounded to the nearest hundredth.

Think of it as a co-ordinate geometry question

the vertices are   (0,0)   (0,8)   (8,0)

Let (r,r) be the centre of the circle

x=0,  y=0  and  y=-x+8  are all tangents to the circle

Points on the circumference are  (r,0), (0,r) (4,4)

Distance from the centre to (0,0) = \(\sqrt{2r^2}=\sqrt2\;r\)

Distance from origin to (4,4) =\( \sqrt{32}=4\sqrt2\)

Distance from origin to (4,4) in terms of raius r is  \( \sqrt2 \;r+r=r(\sqrt2+1)\)

So

\(r(\sqrt2+1)=4\sqrt2\\ r=\frac{4\sqrt2}{\sqrt2+1}=\frac{4\sqrt2(\sqrt2-1)}{1}=8-4\sqrt2=4(2-\sqrt2)\)

Area of circle = \(\pi( 4(2-\sqrt2))^2=16\pi(4+2-4\sqrt2)=32\pi(3-2\sqrt2)\)

Area of triangle = 0.5*8*8 = 32

\(\text{Shaded area}=32-32\pi(3-2\sqrt2)\\ \text{Shaded area}=32(1-3\pi+2\pi\sqrt2)\\ \text{Shaded area}\approx 14.75\; square\; inches\)

You need to check my working.