Graph the image of Kite BCDE after a translation 5 units down.

Umnica [9.8K]

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

(a)/(a^2-16)+(2/(a-4))-(2/(a+4))=0

Simplify ————— a + 4 Equation at the end of step 1 : a 2 2 (—————————+—————)-——— = 0 ((a2)-16) (a-4) a+4 Step 2 : 2 Simplify ————— a - 4 Equation at the end of step 2 : a 2 2 (—————————+———)-——— = 0 ((a2)-16) a-4 a+4 Step 3 : a Simplify ——————— a2 - 16 Trying to factor as a Difference of Squares :

3.1 Factoring: a2 - 16

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =
 A2 - AB + BA - B2 =
 A2 - AB + AB - B2 =
 A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 16 is the square of 4
Check : a2 is the square of a1 

Factorization is : (a + 4) • (a - 4)

Equation at the end of step 3 : a 2 2 (————————————————— + —————) - ————— = 0 (a + 4) • (a - 4) a - 4 a + 4 Step 4 :Calculating the Least Common Multiple :

4.1 Find the Least Common Multiple

 The left denominator is : (a+4) • (a-4)

 The right denominator is : a-4

Number of times each Algebraic Factor
 appears in the factorization of: Algebraic
 Factor Left
Denominator Right
Denominator L.C.M = Max
{Left,Right} a+4 101 a-4 111

Least Common Multiple:
(a+4) • (a-4)

Calculating Multipliers :

4.2 Calculate multipliers for the two fractions

 Denote the Least Common Multiple by L.C.M
 Denote the Left Multiplier by Left_M
 Denote the Right Multiplier by Right_M
 Denote the Left Deniminator by L_Deno
 Denote the Right Multiplier by R_Deno

 Left_M = L.C.M / L_Deno = 1

 Right_M = L.C.M / R_Deno = a+4

Making Equivalent Fractions :

4.3 Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example : 1/2 and 2/4 are equivalent, y/(y+1)2 and (y2+y)/(y+1)3 are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

L. Mult. • L. Num. a —————————————————— = ————————————— L.C.M (a+4) • (a-4) R. Mult. • R. Num. 2 • (a+4) —————————————————— = ————————————— L.C.M (a+4) • (a-4) Adding fractions that have a common denominator :

4.4 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

a + 2 • (a+4) 3a + 8 ————————————— = ————————————————— (a+4) • (a-4) (a + 4) • (a - 4) Equation at the end of step 4 : (3a + 8) 2 ————————————————— - ————— = 0 (a + 4) • (a - 4) a + 4 Step 5 :Calculating the Least Common Multiple :

5.1 Find the Least Common Multiple

 The left denominator is : (a+4) • (a-4)

 The right denominator is : a+4

Number of times each Algebraic Factor
 appears in the factorization of: Algebraic
 Factor Left
Denominator Right
Denominator L.C.M = Max
{Left,Right} a+4 111 a-4 101

Least Common Multiple:
(a+4) • (a-4)

Calculating Multipliers :

5.2 Calculate multipliers for the two fractions

 Denote the Least Common Multiple by L.C.M
 Denote the Left Multiplier by Left_M
 Denote the Right Multiplier by Right_M
 Denote the Left Deniminator by L_Deno
 Denote the Right Multiplier by R_Deno

 Left_M = L.C.M / L_Deno = 1

 Right_M = L.C.M / R_Deno = a-4

Making Equivalent Fractions :

5.3 Rewrite the two fractions into equivalent fractions

L. Mult. • L. Num. (3a+8) —————————————————— = ————————————— L.C.M (a+4) • (a-4) R. Mult. • R. Num. 2 • (a-4) —————————————————— = ————————————— L.C.M (a+4) • (a-4) Adding fractions that have a common denominator :

5.4 Adding up the two equivalent fractions

(3a+8) - (2 • (a-4)) a + 16 ———————————————————— = ————————————————— (a+4) • (a-4) (a + 4) • (a - 4) Equation at the end of step 5 : a + 16 ————————————————— = 0 (a + 4) • (a - 4) Step 6 :When a fraction equals zero : 6.1 When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

a+16 ——————————— • (a+4)•(a-4) = 0 • (a+4)•(a-4) (a+4)•(a-4)

Now, on the left hand side, the (a+4) • (a-4) cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
a+16 = 0

Solving a Single Variable Equation :

6.2 Solve : a+16 = 0

Subtract 16 from both sides of the equation :
 a = -16

One solution was found :

a = -16

4 0

3 years ago

Hope someone can help me with this

RoseWind [281]

(192/16)*4.55*1.43 = 39039/500 =78.078

7 0

4 years ago

Describe the relationship between 2/3 of 3/4 and 3/4 of 2/3?

nexus9112 [7]

The relationship between 2/3 of 3/4 and 3/4 of 2/3 is that the product will equal the same because in multiplication, it doesn't matter what order you put the numbers in because it will equal the same product.

5 0

3 years ago

Plz help, 15 points!

Eduardwww [97]

Answer:

$2x+3y=-6$

Step-by-step explanation:

Let the line is $y=mx+b$ where $m$ is slope and $b$ is $y-intercept$ .

The line is parallel to $2x+3y=3$ , slope of both the lines will be same.

Find the slope of $2x+3y=3$

$3y=3-2x\\y=-\frac{2}{3}x+1$

Slope of line $=-\frac{2}{3}$

$m=-\frac{2}{3}$

So the line will be $y=-\frac{2}{3}x+b$

It passes through $(3,-4)$ .

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

Hence the line is $y=-\frac{2}{3}x-2$

$y=-\frac{2}{3}x-2\\3y=-2x-6\ \ \ \ \ \ \ (Multiply both sides by 3)\\2x+3y=-6$

6 0

3 years ago

A population has the following characteristics. (a) A total of 75% of the population survives the first year. Of that 75%, 25% s

Artemon [7]

Answer:

$= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right] \left \begin{array}{ccc}{0 \ \leq age \leq 1 }\\{ 1 \ \leq age \leq 2 }\\{2 \ \leq age \leq 3}\end{array}\right$

i.e after the first year ;

there 1344 members in the first age class

84 members for the second age class; and

28 members for the third age class

Step-by-step explanation:

We can deduce that the age distribution vector x represents the number of population members for each age class; Given that in each class of age there are 112 members present.

The current age distribution vector is as follows:

$x = \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right] \left[\begin{array}{ccc}{0 \ \leq age \leq 1 }\\{ 0 \ \leq age \leq 2 }\\{0 \ \leq age \leq 3}\end{array}\right]$

Also , the age transition matrix is as follows:

$L = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]$

After 1 year ; the age distribution vector will be :

$x_2 =Lx_1 = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right] \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right]$

$= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right] \left \begin{array}{ccc}{0 \ \leq age \leq 1 }\\{ 1 \ \leq age \leq 2 }\\{2 \ \leq age \leq 3}\end{array}\right$

6 0

3 years ago

Graph the image of Kite BCDE after a translation 5 units down.​

Graph the image of Kite BCDE after a translation 5 units down.