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hammer [34]
2 years ago
11

Two parallel lines are cut by a transversal, as shown below. Solve for b. Then explain your reasoning.

Mathematics
1 answer:
Svetlanka [38]2 years ago
3 0

Answer:

b = 14

Step-by-step explanation:

Since both angles are diagonal each other, they are equal to each other.

5b = b + 56

-b      -b           (subtract b from both sides)

<u>4b</u> = <u>56</u>

4       4            (divide both sides by 4 to solve for b)

 b = 14

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Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a
8_murik_8 [283]

Answer:

  • a) 3/5·((-2)^n + 4·3^n)
  • b) 3·2^n - 5^n
  • c) 3·2^n + 4^n
  • d) 4 - 3 n
  • e) 2 + 3·(-1)^n
  • f) (-3)^n·(3 - 2n)
  • g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form a[n]=r^n, then it will satisfy ...

  r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}

Rearranging and dividing by r^{n-2}, we get the quadratic ...

  r^2-c_1r-c_2=0

The quadratic formula tells us values of r that satisfy this are ...

  r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

  a[n]=pr_1^n+qr_2^n

We can find p and q by solving the initial condition equations:

\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

These have the solution ...

p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}

_____

Using these formulas on the first recurrence relation, we get ...

a)

c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n

__

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

  a[n]=(p+qn)r^n

The initial condition equations are now ...

\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

and the solutions for p and q are ...

p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}

__

Using these formulas on problem (d), we get ...

d)

c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n

__

And for problem (f), we get ...

f)

c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0
3 years ago
Please answer correctly !!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!!!
zimovet [89]

Answer:

For every 2 boys in class, there is 1 girl in class

Step-by-step explanation:

5 0
2 years ago
How does finding the square root of a number compare to finding the cube root lf a number? Use the number 64 in your explainatio
xenn [34]

If you want to know the concept of square roots and cube roots ?

the square root of a number is that number which when multiplied by itself two times gives us that number

e.g.

√64 = x , so that (x)(x) = 64

and from our multiplication table we know that

(8)(8) = 64 , so that

√64 = 8

the cube root of a number is that number which when multiplied by itself three times gives us that number

e.g.

∛64 = x , so that (x)(x)(x) = 64 , and thus x = 4 because 4x4x4 = 64

so ∛64 = 4

6 0
3 years ago
A. -3/2<br><br> b. 3/2<br><br> c. 2/3 <br><br> d. -2/3
zubka84 [21]
D. -2/3 because when you do (y2-y1)/(x2-x1) you get -4/6 which simplifies to -2/3
7 0
3 years ago
Read 2 more answers
Solve for x<br> 8x = y<br> PLEASE HELPPPPPP
svlad2 [7]

Hi ;-)

8x=y \ \ /:8\\\\\boxed{x=\frac{y}{8}}

6 0
3 years ago
Read 2 more answers
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