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Nitella [24]
3 years ago
15

What is the simplest form of /10,000x^64

Mathematics
2 answers:
Akimi4 [234]3 years ago
8 0
Assuming / means square root, we can take the square root of 10,000, which is 100. Then, we can half the exponent on the x, which gets us x^32. Therefore, our answer is 100x^32.

Hope this helps!
nadezda [96]3 years ago
6 0

Answer:

100x^32 just took the test on edge

Step-by-step explanation:

You might be interested in
Which statement is true
loris [4]

Answer:


Step-by-step explanation


You must find the common denominator for all of them.  


A is correct. 13 x 2 is 26 and 14 x 2 is 28. Therefore, 13/14 is greater than 25/28.


B is not correct. 4 x 5 is 20 and 9 x 5 is 45. 4/9 is actually less than 21/45.


C is not correct. 5 x 2 is 10 and 6 x 2 is 12. 5/6 is actually less than 11/12.


D is not correct. 4 x 5 is 20 and 5 x 5 is 25. 4/5 is actually less than 8/25.

4 0
3 years ago
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
2 years ago
Costco is offering to perform the floor
const2013 [10]

Answer:

3.50x= t

Step-by-step explanation:

x square feet of teh floor

t= total cost

3 0
2 years ago
Find the distance between the two points rounding to the nearest tenth
Alika [10]

Answer:

Step-by-step explanation:

The x difference:

8 - 1 = 7

The y difference:

-4 - -7 = 3

Distance:

7² + 3² = 58

√58 ≅ 7.62

Please Thank and Mark as Brainliest!

3 0
2 years ago
Which of these numbers is four less than a square number?<br>A 3<br>B 20<br>C 37<br>D 45​
gregori [183]

Answer:

D

Step-by-step explanation:

You first add 4 to all of the numbers giving you 7, 24, 41, and 49. Only choice D is a perfect square. for reference, 7*7=49

4 0
2 years ago
Read 2 more answers
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