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kherson [118]
2 years ago
15

Damian reads 21 pages in 1 hour. How many pages can he read in 3 hours?

Mathematics
2 answers:
irinina [24]2 years ago
3 0
If he can read 21 pages in 1 hour he can read 63 in 3 hours
Stolb23 [73]2 years ago
3 0

Answer:

If Damian reads an average of 21 pages an hour, he should be able to read 63 pages in 3 hours.

Step-by-step explanation:

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Elizabeth’s aunt started a college savings account for her with $6,000. What is the balance of the account after 5 years if the
vaieri [72.5K]

Answer:

6750

Step-by-step explanation:

simple interest = P×r%×t

P = $6000

r = 2.5%

t = 5 years

I = 6000 × 2.5/100 × 5

= 750

6000 + 750 = $6750

7 0
2 years ago
Helpppppp!!!!!!!! Please
stiv31 [10]

Hey there!!

We know angle DEC = angle AEB

Hence,

... 4x + 12 = 5x - 10

... - x = - 22

... x = 22

4 ( 22 ) + 12

88 + 12

100° is the answer.

Hope my answer helps!

7 0
3 years ago
Ratchpad<br> Which line of reflection maps point Q at (1, 2) to point Q at (-2,-1)?
allochka39001 [22]
Idk...................................
7 0
3 years ago
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
The price of a cellular telephone plan is based on peak and nonpeak service. Kelsey used 45 peak minutes and 50 nonpeak minutes
Oxana [17]
We can build a system of two linear equations with two unknowns with the info provided in the problem, one with Kelsey info and one with Mitch info like so:
Lets call p the amount on peak minutes and n the amount of non-peak minutes:
45p + 50n = 27.75
70p + 30n = 36
lets reduce the equations dividing the first by 5:
9p + 10n = 5.55
<span>70p + 30n = 36
</span>now, to eliminate n, lets multiply the first equation by -3 and add the two equations:
-27p - 30n = -16.65
<span>70p + 30n = 36
</span>----------------------------
43p + 0 = 19.35
p = 19.35<span>/43
p = 0.45
therefore the peak rate is $0.45 per minute
lets substitute in one of the original equations this result:
</span><span>45p + 50n = 27.75
</span>45(0.45) + 50n = 27.75
20.25 <span>+ 50n = 27.75
50n = 27.75 - 20.25
50n = 7.5
n = 7.5/50
n = 0.15
therefore the non-peak rate per minute is $0.15</span>
5 0
2 years ago
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