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stealth61 [152]
3 years ago
10

What are the solutions to this equation?

Mathematics
1 answer:
prisoha [69]3 years ago
6 0

Answer:

x = ±1

Step-by-step explanation:

16x² = 16

x² = 1

x = ±1

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The blue jay path at the state park goes around the perimeter of the park as shown in the map below
diamong [38]
What map?? what path???? wheres the blue jay?!?!
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3 years ago
8 is 4% of the number
Murljashka [212]

Answer:

(8/4)100 = 200

Step-by-step explanation:

Simplify the following:

8/4×100

Hint: | Express 8/4×100 as a single fraction.

8/4×100 = (8×100)/4:

(8×100)/4

Hint: | In (8×100)/4, divide 100 in the numerator by 4 in the denominator.

4 | | 2 | 5

| 1 | 0 | 0

- | | 8 |  

| | 2 | 0

| - | 2 | 0

| | | 0:

8×25

Hint: | Multiply 8 and 25 together.

8×25 = 200:

Answer: 200

5 0
3 years ago
Read 2 more answers
Approximate the stationary matrix S for the transition matrix P by computing powers of the transition matrix P.
Scrat [10]

Answer:

S = [0.2069,0.7931]

Step-by-step explanation:

Transition Matrix:

P=\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

Stationary matrix S for the transition matrix P is obtained by computing powers of the transition matrix P ( k powers ) until all the two rows of transition matrix p are equal or identical.

Transition matrix P raised to the power 2 (at k = 2)

P^{2} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{2} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right]

Transition matrix P raised to the power 3 (at k = 3)

P^{3} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{3} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

  P^{3} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right]

Transition matrix P raised to the power 4 (at k = 4)

P^{4} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{4} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{4} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right]

Transition matrix P raised to the power 5 (at k = 5)

P^{5} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{5} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]

P^{5} =\left[\begin{array}{ccc}0.2069&0.7931\\0.2069&0.7931\end{array}\right]

P⁵ at k = 5 both the rows identical. Hence the stationary matrix S is:

S = [ 0.2069 , 0.7931 ]

6 0
4 years ago
A scientist has 100 milligrams of a radioactive element. The amount of radioactive element remaining after t days can be determi
pav-90 [236]

The answer is B. 81.2%

8 0
3 years ago
This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex
olga nikolaevna [1]

Answer:

maximum value , f max = 169 (taking the most out of the 4th power)

minimum value , f min = 169/3 (taking the least out of the 4th power)

Step-by-step explanation:

A) Since both function and restriction are symmetrical with respect to x,y and z, there is no reason for one to be more important than the others and therefore one solution would be x=y=z=λ and thus

x2 + y2 + z2 = 13  → 3λ² = 13 →λ² = 13/3

and f would be

f (x, y, z) = x4 + y4 + z4 = 3λ⁴ = 3*(13/3)²=13²/3=169/3

since x⁴ increases faster than 3*x² , f(x,y,z) would be a minimum

and the maximum value would be obtained taking the most out of x⁴, thus doing 2 coordinates =0 ( can be x=0 and y=0) and

z²= 13

f (x, y, z) = x4 + y4 + z4 = 13² = 169

B) strictly, using Lagrange multipliers

f (x, y, z) = x4 + y4 + z4

g (x, y, z) = x2 + y2 + z2 - 13

F(x,y,z) = f (x, y, z) -λ*g (x, y, z)

such that

Fx (x,y,z)=  fx(x, y, z) -λ*gx (x, y, z) = 0 → 4*x³ - λ*2*x = 0 → 2*x*(2*x² -λ) = 0

thus x=0 or x²= λ/2

Fy (x,y,z)=  fy(x, y, z) -λ*gy (x, y, z)= 0 → 4*y³ - λ*2*y = 0 → 2*y*(2*y² - λ) = 0

thus y=0 or y²= λ/2

Fz (x,y,z)=  fz(x, y, z) -λ*gz (x, y, z)= 0 → 4*z³ - λ*2*z = 0→ 2*z*(2*z² - λ) = 0

thus z=0 or z²= λ/2

g (x, y, z) = 0  → x2 + y2 + z2 = 13 → 3*(λ/2) = 13 → λ=13*2/3

thus  x²=y²=z²= λ/2 =13/3

f min = f (x, y, z) = x4 + y4 + z4 = 3*(13/3)²=169/3

for the x=0 , y=0 → z²= 13

f max = f (x, y, z) = x4 + y4 + z4 = 13² = 169

3 0
3 years ago
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