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LuckyWell [14K]

3 years ago

7

The standard form of 230 000

2 answers:

Lisa [10]3 years ago

6 0

200,000 + 30,000 or two hundred thirty thousand I believe. hope this helps

nikitadnepr [17]3 years ago

3 0

230,000
English = Two hundred thousand and thirty thousand.

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Renna pushes the elevator button, but the elevator does not move. The mass limit for the elevator is

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Answer:

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3 years ago

If x-1/x=5 find x^3+1/x^3=?

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$\dfrac{x-1}{x}=5\ \ \ \ /x\neq0\\\\\dfrac{x-1}{x}=\dfrac{5}{1}\ \ \ \ |\text{cross multiply}\\\\5x=x-1\ \ \ \ |-x\\\\5x=1\ \ \ \ |:5\\\\x=0.2\\\\\dfrac{x^3+1}{x^3}=\dfrac{x^3}{x^3}+\dfrac{1}{x^3}=1+\dfrac{1}{x^3}\to1+\dfrac{1}{0.2^3}=1+\dfrac{1}{0.008}\\\\=1+\dfrac{1}{\frac{8}{1,000}}=1+\dfrac{1,000}{8}=1+125=126\\\\\text{Answer:}\ \dfrac{x^3+1}{x^3}=126$

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4 years ago

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4 years ago

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Step-by-step explanation:

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3 years ago

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kvv77 [185]

Answer: Option C

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Step-by-step explanation:

Whenever we have a main function f(x) and we want to transform the graph of f(x) by moving it vertically then we apply the transformation:

$k (x) = f (x) + b$

If $b> 0$ then the graph of k(x) will be the graph of f(x) displaced vertically b units down.

If $b> 0$ then the graph of k(x) will be the graph of f(x) displaced vertically b units upwards.

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Then, to move its vertex 7 units down we apply the transformation:

$k (x) = f (x) - 7\\\\k (x) = x ^ 2 -7$ .

Then the function k(x) that will have its vertex 7 units below f(x) is

$k (x) = x ^ 2 -7$

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3 years ago