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3 years ago

11

Idk how to solve 0=4+n over 5

1 answer:

miv72 [106K]3 years ago

5 0

<span> 0=4+n over 5
In number form the equation is
</span> $0= \frac{4+n}5}$

<span>We want to isolate n, so to undue division we multiply, we'll multiply each side by 5 Leaving the equation as
</span> $0=20+n$
<span>
Next we'll subtract 20 from each side to isolate n
The equation would now be
</span> $-20=n$
<span>
So the answer is N=-20

</span>

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9. Let f be a continuous function such that f(3) = 4 and f(5)=2. Let h(x)= f(x) + 2x. Explain why

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Since <em>f(x)</em> is continuous, so is <em>h(x)</em>.

Also, since <em>f</em> (3) = 4 and <em>f</em> (5) = 2, we have

<em>h</em> (3) = <em>f</em> (3) + 2•3 = 4 + 6 = 10

<em>h</em> (5) = <em>f</em> (5) + 2•5 = 2 + 10 = 12

It follows from the intermediate value theorem that there is some <em>t</em> between 3 and 5 such that <em>h(t)</em> is between 10 and 12, and namely there is some <em>t</em> for which <em>h(t)</em> = 11.

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3 years ago

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3 years ago

Read 2 more answers

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soldier1979 [14.2K]

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2 years ago

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2 years ago

Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val

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Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

$A(t) = Pe^{rt}$

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

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This means that $P = 330$

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

$A(t) = Pe^{rt}$

$590 = 330e^{8r}$

$e^{8r} = \frac{590}{330}$

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Applying ln to both sides:

$\ln{e^{8r}} = \ln{1.79}$

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4 years ago