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3 years ago

Which number produces a rational number when multiplied by 1/2

Mathematics

2 answers:

Bogdan [553]3 years ago

5 0

Step-by-step explanation:

all odd numbers

that should be the answer

Ulleksa [173]3 years ago

3 0

Answer:

The other number should also be rational

so lets say one of your answer choices is 0.314 then that would be your answer.

Step-by-step explanation:

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Gwen takes out a loan of $400 to pay for an online course. Flat rate interest is charged at 8% p.a

puteri [66]

We know, 1 year = 12 month.

So, 3 month = 3/12 = 0.25 year.

Now, principle amount, P = $400 .

Rate is, r = 8% = 0.08

Time period, t = 0.25 year.

Now, Interest is given by :

I = P × r × t

I = $( 400 × 0.08 × 0.25 )

I = $8

Therefore, she will pay $8 interest in total.

5 0

3 years ago

Choose the graph of the solution to this inequality.

Dmitriy789 [7]

Answer:

we can translate the given statement into-8 ≥ -5x + 2 > -38.

In this case, we can dissect the inequality into two parts:

-8 ≥ -5x + 2 and -5x + 2 > -38

Step-by-step explanation:

-8 ≥ -5x + 2 5x ≥ 10x ≥ 2 (closed dot)

-5x + 2 > -38 5x < 40x < 8 (open dot)

The answer then is c) number line with a closed dot on 2 and an open dot on 8 and shading in between

8 0

2 years ago

Evaluate the limits

julsineya [31]

$x > \ln(x)$ for all $x$ , so

$\displaystyle \lim_{x\to\infty} (\ln(x) - x) = - \lim_{x\to\infty} x = \boxed{-\infty}$

Similarly, $\displaystyle \lim_{x\to\infty} (x-e^x) = - \lim_{x\to\infty} e^x = \boxed{-\infty}$

We can of course see the limits are identical by replacing $x\mapsto e^x$ , so that

$\displaystyle \lim_{x\to\infty} (\ln(x) - x) = \lim_{x\to\infty} (\ln(e^x) - e^x) = \lim_{x\to\infty} (x - e^x)$

You can also rewrite the limands to accommodate the application of l'Hôpital's rule. For instance,

$\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\exp\left(\lim_{x\to\infty} (x - e^x)\right)\right) = \ln\left(\lim_{x\to\infty} e^{x-e^x}\right) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right)$

Using the rule, the limit here is

$\displaystyle \lim_{x\to\infty} \frac{(e^x)'}{\left(e^{e^x}\right)'} = \lim_{x\to\infty} \frac{e^x}{e^x e^{e^x}} = \lim_{x\to\infty} \frac1{e^{e^x}} = 0$