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irinina [24]
2 years ago
9

1" title="\left(\sqrt{3}+4\right)\left(1+\sqrt{3}\right)" alt="\left(\sqrt{3}+4\right)\left(1+\sqrt{3}\right)" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
kirza4 [7]2 years ago
4 0

Answer:

\huge\boxed{\sf 7 + 5\sqrt{3} }

Step-by-step explanation:

(\sqrt{3} +4)(1+\sqrt{3})\\\\= \sqrt{3}(1+\sqrt{3} )+4(1+\sqrt{3})\\\\= \sqrt{3} + (\sqrt{3} )^2 + 4 + 4\sqrt{3} \\\\= \sqrt{3} + 3+4+4\sqrt{3}  \\\\= 7 + \sqrt{3}  + 4\sqrt{3} \\\\Take \ \sqrt{3} \ common\\\\= 7 + \sqrt{3} (1+4)\\\\= 7 + \sqrt{3}(5)\\\\= 7 + 5\sqrt{3} \\\\\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
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Jay makes a base salary of $2000 per month. Once he reaches $50000 in total sales, he earns an additional 5% commission (salary)
ki77a [65]

Answer:

j(s) = 0.05s -500

Step-by-step explanation:

Given

Base\ Salary = 2000

Commission = 5\% on Sales over 50000

Required

Determine the function j(s) for sales over 50000

Represent the total sales in a month with s.

Sales over 50000 in that month will be: s - 50000

So, the function j(s) is:

j(s) = Base Amount + Commission * Sales over 50000

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Convert % to decimal

j(s) = 2000 + 0.05 * (s - 50000)

Open bracket

j(s) = 2000 + 0.05 * s - 0.05 * 50000

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Collect Like Terms

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3 0
3 years ago
What is the domain of F(x) = In(x)?
jekas [21]

Answer as an inequality:  x > 0

Answer in interval notation:  (0, \infty)

Answer in words: Set of positive real numbers

All three represent the same idea, but in different forms.

======================================================

Explanation:

Any log is the inverse of an exponential equation. Consider a general base b such that f(x) = b^x. The inverse of this is f^{-1}(x) = \log_b(x)

For the exponential b^x, we cannot have b^x = 0. We can get closer to it, but we can't actually get there. The horizontal asymptote is y = 0.

Because of this, \log_b(x) has a vertical asymptote x = 0 (recall that x and y swap, so the asymptotes swap as well). This means we can get closer and closer to x = 0 from the positive side, but never reach x = 0 itself.

The domain of \log_b(x) is x > 0 which in interval notation would be (0, \infty). This is the interval from 0 to infinity, excluding both endpoints.

------------------------

The natural log function Ln(x) is a special type of log function where the base is b = e = 2.718 approximately.

So,

\log_e(x) = \text{Ln}(x)

allowing all of what was discussed in the previous section to apply to this Ln(x) function as well.

------------------------

In short, the domain is the set of positive real numbers. We can't have x be 0 or negative.

8 0
3 years ago
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