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3 years ago

10 x 3 thousands in standard form

Mathematics

2 answers:

Serjik [45]3 years ago

8 0

Answer:

30,000

Step-by-step explanation:

valina [46]3 years ago

7 0

The answer is 30,000

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Find the absolute maximum and minimum values of the function below. f(x) = x3 − 9x2 + 3, − 3 2 ≤ x ≤ 12 Solution Since f is cont

Neko [114]

Answer:

There are an absolute minimum (x = 6) and an absolute maximum (x = 12).

Step-by-step explanation:

The correct statement is described below:

Find the absolute maximum and minimum values of the function below:

$f(x) = x^{3}-9\cdot x^{2}+ 3$ , $2 \leq x \leq 12$

Given that function is a polynomial, then we have the guarantee that function is continuous and differentiable and we can use First and Second Derivative Tests.

First, we obtain the first derivative of the function and equalize it to zero:

$f'(x) = 3\cdot x^{2}-18\cdot x$

$3\cdot x^{2}-18\cdot x = 0$

$3\cdot x \cdot (x-6) = 0$ (Eq. 1)

As we can see, only a solution is a valid critical value. That is: $x = 6$

Second, we determine the second derivative formula and evaluate it at the only critical point:

$f''(x) = 6\cdot x -18$ (Eq. 2)

x = 6

$f''(6) = 6\cdot (6)-18$

$f''(6) =18$ (Absolute minimum)

Third, we evaluate the function at each extreme of the given interval and the critical point as well:

x = 2

$f(2) = 2^{3}-9\cdot (2)^{2}+3$

$f(2) = -25$

x = 6

$f(6) = 6^{3}-9\cdot (6)^{2}+3$

$f(6) = -105$

x = 12

$f(12) = 12^{3}-9\cdot (12)^{2}+3$

$f(12) = 435$

There are an absolute minimum (x = 6) and an absolute maximum (x = 12).

6 0

3 years ago

What is 13x= 2x+26-12.5?

Ulleksa [173]

The answer would be 6, 0.5

5 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 ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

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 ...

$a[n]=(p+qn)r^n$