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Olin [163]
3 years ago
11

How do you put 943,261,586 with base ten numbers

Mathematics
1 answer:
Citrus2011 [14]3 years ago
8 0

Answer:

(9 \times 10^{8}) + (4 \times 10^{7}) + (3 \times 10^{6}) + (2 \times 10^{5}) +  (6 \times 10^{4}) +  (10^{3}) +  (5 \times 10^{2}) +  (8 \times 10) + (6 \times 10^{0})

Step-by-step explanation:

How do you put 943,261,586 with base ten numbers

943,261,586  =  

900,000,000 + 40,000,000 + 3,000,000 + 200,000 + 60,000 + 1,000 + 500 + 80 +6 =

(9 \times 10^{8}) + (4 \times 10^{7}) + (3 \times 10^{6}) + (2 \times 10^{5}) +  (6 \times 10^{4}) +  (10^{3}) +  (5 \times 10^{2}) +  (8 \times 10) + (6 \times 10^{0})

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ANSWER:

\:D.\text{ }f^{-1}\left(x\right)=\left(x+4\right)^2;x\ge-4

STEP-BY-STEP EXPLANATION:

We have the following equation:

f(x)=\sqrt{x}-4

The inverse is the following (we calculate it by replacing f(x) by x and x by f(x)):

\begin{gathered} x=\sqrt{f^{-1}(x)}-4 \\  \\ \sqrt{f^{-1}(x)}=x+4 \\  \\ f^{-1}(x)=(x+4)^2 \end{gathered}

The domain would be the range of the original equation, and it would be the range of values that f(x) could take, which was from -4 to positive infinity, that is, f(x) ≥ -4.

Therefore, the domain is x ≥ -4.

So the correct answer is D.

\:f^{-1}\left(x\right)=\left(x+4\right)^2;x\ge -4

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The 3rd term of (a-b)4 is
fgiga [73]

The third term of the expansion is 6a^2b^2

<h3>How to determine the third term of the expansion?</h3>

The binomial term is given as

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So, we have

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Hence, the third term of the expansion is 6a^2b^2

Read more about binomial expansion at

brainly.com/question/13602562

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