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galben [10]
2 years ago
8

The product of three consecutive numbers is 21,924. What is the smallest number out these numbers?

Mathematics
1 answer:
Natasha_Volkova [10]2 years ago
7 0

Answer:

27

Step-by-step explanation:

Let the middle number be x.

(x - 1)(x)(x + 1) = 21,924

You can solve that equation, or you can use logic.

The 3 numbers are going to be close to the cubic root of 21,924.

cubic root of 21,924 = 27.988 or approx 28.

Try 27, 28, 29

27 * 28 * 29 = 21,924

Answer: 27

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5 0
3 years ago
I need help asap T>T
n200080 [17]

Answer:

3.4 yards cubed

Step-by-step explanation:

Volume is length x width x height

Length 2 4/5, Width 2, Height 3/5 so 2 4/5 x 2 x 3/5

It looks like they want the answer in decimal form, so you can convert the fractions into decimals.

For the length 2 4/5,  you divide the 4 by 5 to get a decimal of .8 so now it's 2.8

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Now it's 2.8 x 2 x .6 = 3.36 yards cubed which could get rounded up to 3.4 yards cubed

6 0
2 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
3 years ago
What is the area of a circle that has a diameter of 18"?
lubasha [3.4K]
Since the formula for the area of a circle is Pi x r squared, the area would be 81 x Pi (you teacher may want you to use Pi as 3.14 or put it in your calculator).
7 0
3 years ago
Read 2 more answers
The solution to the system is (6,2) Select all the equations that could be the other equation in the system. PLZ HELPPPPPP there
Dovator [93]

9514 1404 393

Answer:

  A, C

Step-by-step explanation:

The attached graph shows which lines go through the given point. They are ...

  y = 1/2x -1 . . . . 1st selection

  y = -1/6x +3 . . . 3rd selection

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The equations can be found algebraically by substituting the given point in the equation and seeing if the result is a true statement.

  a) 2 = (1/2)(6) -1 = 3 -1 . . . true

  b) 2 = -3(6) . . . . false

  c) 2 = -1/6(6) +3 = -1 +3 . . . true

  d) 2 = 2/3(6) -1 = 4 -1 . . . . false

  e) 2 = 4(6) -2 = 24 -2 . . . . false

  f) 2 = -3/2(6) +6 = -9 +6 . . . . false

6 0
2 years ago
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