1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
natta225 [31]
2 years ago
11

PLEASE HELP!!! (Thank you) :)

Mathematics
2 answers:
WINSTONCH [101]2 years ago
8 0
The answer is 26 divided by 2 witch is 13
san4es73 [151]2 years ago
7 0

Answer:

13

Step-by-step explanation:

(24 + 2) / 2

26/2

=13

You might be interested in
Please help need answer
Ludmilka [50]

Answer:

A^2 + B^2 = C^2

4 0
3 years ago
Read 2 more answers
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
-4 + 4d + 17 = 33<br> D = ?
cricket20 [7]

Answer:

3?

Step-by-step explanation:

I think it is 3, because...

33 - 17 = 16

16 - 4 = 12

12/4 = 3

???

7 0
2 years ago
Do you know What is X=9+y-2
Damm [24]

Answer:

Subtract

2

from

9

.

X

=

y

+

7

Step-by-step explanation:

5 0
3 years ago
Which equation does the graph represent?
Art [367]

Answer:

It is the second answer

Step-by-step explanation:

The standard form of an ellipse is

x^2/a^2 + y^2/b^2  or x^2/b^2 + y^2/a^2 = 1

If the x is the main axis we use the first form.  If the y is the main axis we use the second form.  We will use the second form.

our a is 3 and our b is 4

x^2/3^2 + y^2/4^2

3 0
2 years ago
Other questions:
  • Line A and Line B are parallel lines cut by transversals x and y. If 2 = 57 and 3 = 85, find the measure of 5
    5·1 answer
  • In 1980, wind turbines in Europe generated about 5 gigawatt-hours of energy. Over the next 15 years, the amount of energy increa
    12·1 answer
  • current Law requires that wheelchair ramps have a height to length ratio of 1: 12 and old ramp has a height to length ratio of 2
    13·1 answer
  • The lifetime of a leaf blower is exponentially distributed with a mean of 5 years. If you buy a 12 year old leaf blower, what is
    7·1 answer
  • Which equation could represent the relationship shown in the scatter plot?
    7·1 answer
  • Guyyyy Please help I have no idea what the answer is
    9·2 answers
  • HELP PLEASE!!
    13·1 answer
  • Which survey question is biased?
    15·2 answers
  • Can someone please determine this for me
    5·1 answer
  • Help help help plsssss and tyyy
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!