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3 years ago

Select the correct answer from each drop-down menu.

2 answers:

6 0

I believe 1 is 3.8, b/c 1.9*2 is 3.8, and the difference between exponents makes 3 zeros, so 2000.
The second one is 40, I confirmed this by doing .0001 to replicate 10 to the negative 4th power
By doing 9.5*.0001*40, I got the 0.038 from 1

Kisachek [45]3 years ago

3 0

Answer:

Well, 1 is 3.8, cause 1.9*2 is 3.8, and the difference between exponents makes 3 0's, so 2000.

The second one is 40, I got this by doing .0001 to replicate 10 to the negative 4th power

when doing 9.5*.0001*40, I got the 0.038 from 1

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Of 560 marbles in a bag, 65% are red and the rest are blue. After 28 red marbles are replaced with blue ones, how many blue marb

jonny [76]

Answer: The answer is 400 blue marbles.

Step-by-step explanation: Given that there are 560 marbles in a bag, out of which 65% are red and rest are blue.

So, number of red marbles is

$n_r=65\%\times 560=\dfrac{65}{100}\times 560=364,$

and number of blue marbles is

$n_b=560-364=196.$

Now, if 28 red marbles are replaced by blue marbles, the the new number of red and blue marbles will be

$n_r^\prime=364-28=336~~~\textup{and}~~~n_b^\prime=196+28=224.$

Now, to get 65% of the marbles blue, we need to add some more blue marbles to the bag. Let 'x' number of blue marbles are added to the bag, then

$65\%\times (560+x)=224+x\\\\\Rightarrow \dfrac{65}{100}\times (560+x)=224+x\\\\\Rightarrow 13(560+x)=20(224+x)\\\\\Rightarrow 7280+13x=4480+20x\\\\\Rightarrow 7x=2800\\\\\Rightarrow x=400.$

Thus, 400 blue marbles need to be added to the bag.

5 0

3 years ago

Read 2 more answers

Help Please I did to finsh

ki77a [65]

Step-by-step explanation:

Hello there!

Again, plug this into the Pythagorean Theorem.

$a^2+b^2=c^2\\2^2+b^2=6^2\\4+b^2=36\\b^2=32\\b=5.65$

7 0

3 years ago

Read 2 more answers

For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl

nevsk [136]

I'm assuming you're talking about the indefinite integral

$\displaystyle\int e^{-x^2/2}\,\mathrm dx$

and that your question is whether the substitution $u=\dfrac x{\sqrt2}$ would work. Well, let's check it out:

$u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}$
$\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du$
$=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du$

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried $u=\sqrt t$ next? Then $\mathrm du=\dfrac{\mathrm dt}{2\sqrt t}$ , giving

$=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt$

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

$\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt$

By the fundamental theorem of calculus, taking the derivative of both sides yields

$\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}$

and so the antiderivative would be

$\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)$

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.

3 0

3 years ago

NEED HELP PLEASE!!<br><br> 3 to the power of 4 + 2 ⋅ 5 = ____. (Input whole numbers only.)

olya-2409 [2.1K]

3^4 + 2 * 5 = 91

The answer is 91.

6 0

3 years ago

Please help me out ASAP

Nady [450]

Answer:

the answer is b

Step-by-step explanation:

3 0

3 years ago