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Gennadij [26K]
3 years ago
8

If ∠a and ∠b are supplementary and m∠a = 37°45', then m∠b =

Mathematics
1 answer:
dlinn [17]3 years ago
5 0

a + b = 180

37.45 + b = 180

b = 180-37.45 = 142.55

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Find the general term of sequence defined by these conditions.
disa [49]

Answer:

\displaystyle  a_{n}  =     (2)^{2n -1}   -   (3) ^{n-1 }

Step-by-step explanation:

we want to figure out the general term of the following recurrence relation

\displaystyle \rm a_{n + 2} - 7a_{n + 1} + 12a_n = 0  \:  \: where :  \:  \:a_1 = 1 \: ,a_2 = 5,

we are given a linear homogeneous recurrence relation which degree is 2. In order to find the general term ,we need to make it a characteristic equation i.e

  • {x}^{n}  =  c_{1} {x}^{n - 1}  + c_{2} {x}^{n - 2}  + c_{3} {x}^{n -3 } { \dots} + c_{k} {x}^{n - k}

the steps for solving a linear homogeneous recurrence relation are as follows:

  1. Create the characteristic equation by moving every term to the left-hand side, set equal to zero.
  2. Solve the polynomial by factoring or the quadratic formula.
  3. Determine the form for each solution: distinct roots, repeated roots, or complex roots.
  4. Use initial conditions to find coefficients using systems of equations or matrices.

Step-1:Create the characteristic equation

{x}^{2}  - 7x+ 12= 0

Step-2:Solve the polynomial by factoring

factor the quadratic:

( {x}^{}  - 4)(x - 3) =  0

solve for x:

x =  \rm 4 \:and \: 3

Step-3:Determine the form for each solution

since we've two distinct roots,we'd utilize the following formula:

\displaystyle a_{n}  = c_{1}  {x} _{1} ^{n }  + c_{2}  {x} _{2} ^{n }

so substitute the roots we got:

\displaystyle a_{n}  = c_{1}  (4)^{n }  + c_{2}  (3) ^{n }

Step-4:Use initial conditions to find coefficients using systems of equations

create the system of equation:

\begin{cases}\displaystyle 4c_{1}    +3 c_{2}    = 1  \\ 16c_{1}    + 9c_{2}     =  5\end{cases}

solve the system of equation which yields:

\displaystyle c_{1}  =  \frac{1}{2}     \\  c_{2}   =   - \frac{1}{3}

finally substitute:

\displaystyle  a_{n}  =  \frac{1}{2}   (4)^{n }   -  \frac{1}{3}  (3) ^{n }

\displaystyle \boxed{ a_{n}  =    (2)^{2n-1 }   -   (3) ^{n -1}}

and we're done!

7 0
3 years ago
Mrs. Daniel packs reams of paper into boxes. The box has a mass of 18
melisa1 [442]
Should be B the total numbers of boxes filled with reams
7 0
3 years ago
240,000 written as 10th power
kirill115 [55]
The answer is 24 to the 4th power
3 0
3 years ago
Determine the x and y intercepts for the line 3x – 5y + 15 = 0. Show your work.
iragen [17]

Answer:

Step-by-step explanation:

The x and y intercepts occur when either x or y = 0

For the y intercept, x = 0

3(0) - 5y + 15 = 0

- 5y + 15 = 0                  Subtract 15 from both sides.

-5y = - 15                       Divide by - 5

-5y / -5 = - 15/-5

y = 3

For x intercept, y = 0

3x - 5(0) + 15 = 0

3x + 15 = 0                    Subtract 15 from both sides

3x = - 15                        Divide by 3

3x/3 = - 15/3

x = - 5

xintercept = (-5,0)

yintercept = (0,3)

3 0
2 years ago
Solve for x<br><br> ax - k = 3(x + h)
inysia [295]

Answer:

x=3h+k/a-3

Step-by-step explanation:

Attached is the solution. I hope this helps you!!

8 0
3 years ago
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