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3 years ago

5

Round 32,521 to the nearest ten thousand

2 answers:

joja [24]3 years ago

4 0

When u round 32,521 to the nearest ten it would be 30,000

Rom4ik [11]3 years ago

3 0

When you round 32,521 to the nearest ten thousand, it rounds to 32,000.

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For the function h(x) defined below, find function f(x) and g(x) so that h(x)=(f^ g)(x) . h(x) = ((x + 2)/x) ^ 3

dem82 [27]

Answer:

Step-by-step explanation:

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3 years ago

What is the solution to sqrt 17-x=x+3? Show your work.

snow_lady [41]

Answer:

$x=1$

Step-by-step explanation:

Remember:

$(\sqrt[n]{a})^n=a\\\\(a+b)=a^2+2ab+b^2$

Given the equation $\sqrt{17-x}=x+3$ , you need to solve for the variable "x" to find its value.

You need to square both sides of the equation:

$(\sqrt{17-x})^2=(x+3)^2$

$17-x=(x+3)^2$

Simplifying, you get:

$17-x=x^2+2(x)(3)+3^2\\\\17-x=x^2+6x+9\\\\x^2+6x+9+x-17=0\\\\x^2+7x-8=0$

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$(x-1)(x+8)=0$

Then:

$x_1=1\\x_2=-8$

Let's check if the first solution is correct:

$\sqrt{17-(1)}=(1)+3$

$4=4$ (It checks)

Let's check if the second solution is correct:

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Therefore, the solution is:

$x=1$

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3 years ago

Solve the following recurrence relation: <br> <img src="https://tex.z-dn.net/?f=A_%7Bn%7D%3Da_%7Bn-1%7D%2Bn%3B%20a_%7B1%7D%20%3D

-Dominant- [34]

By iteratively substituting, we have

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and the pattern continues down to the first term $a_1=0$ ,

$a_n = a_{n - (n - 1)} + n + (n - 1) + (n - 2) + \cdots + (n - (n - 2))$

$\implies a_n = a_1 + \displaystyle \sum_{k=0}^{n-2} (n - k)$

$\implies a_n = \displaystyle n \sum_{k=0}^{n-2} 1 - \sum_{k=0}^{n-2} k$

Recall the formulas

$\displaystyle \sum_{n=1}^N 1 = N$

$\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2$

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$\implies a_n = \dfrac12 n^2 + \dfrac12 n - 1$

$\implies \boxed{a_n = \dfrac{(n+2)(n-1)}2}$

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borishaifa [10]

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Leto [7]

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4 years ago

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