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SSSSS [86.1K]
3 years ago
12

h.) How many different ways can Johnny place an algebra book, a geometry book, a chemistry book, an English book, and a health b

ook on a shelf?​
Mathematics
1 answer:
Nady [450]3 years ago
7 0

Answer:

5!

Step-by-step explanation:

take algebra book. it can be placed first or 2nd or 3rd or 4th or last. so 5 different ways. but geometry book will be arranged in 4 ways, since algebra book is already arranged.

5p5

5!/0!= 5! = 120

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Mhanifa please help fast
Shkiper50 [21]

Answer:

<h3>Q 17</h3>

a is supplementary with 36 and b is supplementary with 113.

  • a = 180 - 36 = 144
  • b = 180 - 113 = 67

Correct choice is D

<h3>Q 2</h3>
  • LM = 1/2(AB + DC) = 1/2(46 + 125) = 85.5
<h3>Q 3</h3>

Diagonals of a rectangle are congruent

  • KM = LN
  • 6x + 16 = 49
  • 6x = 33
  • x = 33/6
  • x = 5.5

Correct choice is A

8 0
3 years ago
What is Obama's last name?
Vinvika [58]

Answer:

OBAMA

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Sue ran 27 miles before she blistered her heel, and then she walked 1 additional mile. Her running speed was 18 times as fast as
natulia [17]

Answer:

Well I think either you already solved this or you forgot to put down the question.... Stay safe and have a great weekend... :)

Step-by-step explanation:

4 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
A) x³y²if x = 2 and y = 4
kow [346]

Answer:

128

Step-by-step explanation:

x³y²if x = 2 and y = 4

x³*y²

(2)^{3} * (4)^{2}

(2*2*2)*(4*4)

(8)*(16)

128

8 0
2 years ago
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