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Trava [24]
2 years ago
5

Is the inequality true for all values of x? (3x+8)^2 > 3x(x+16)

Mathematics
1 answer:
emmainna [20.7K]2 years ago
6 0

Answer:

Yes

Step-by-step explanation:

yes

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Kunal told Ana he was going to drop 10 dimes on the table. He said he would give her all the dimes where the "head" side was sho
sweet-ann [11.9K]

Answer:

10

Step-by-step explanation:

4 0
2 years ago
Karl set out to Alaska on his truck.
tester [92]

Answer:

<u>The track consumes ≅ 61 liters of fuel every 100 kilometres</u>

Step-by-step explanation:

As we can see in the graph, the total distance that the truck can travel with 500 liters of fuel is ≅ 825 kilometres.

For answering the question properly, we use the Rule of Three Simple, this way:

Kilometres                Liters of fuel

825                              500

100                                 x

Solving for x, we have:

825 * x = 500 * 100

825x = 50,000

x = 50,000/825

x = 60.6 liters of fuel (61 rounding to the next whole)

x ≅ 61 liters of fuel

<u>The track consumes ≅ 61 liters of fuel every 100 kilometres</u>

6 0
3 years ago
Read 2 more answers
PLEASE HELP!!!<br> (9r^3 - 59r^2 - 30r + 7) / (r - 7)
Xelga [282]
Go to math papa it really helps

8 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
Write 645,300 in expanded form using exponents
Gemiola [76]

Answer:

Step-by-step explanation:

(6 x 10^5) + (4 x 10^4) + (5 x 10^3) + (3 x 10^2)

6 0
3 years ago
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