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3 years ago

2^5×8^4/16=2^5×(2^a)4/2^4=2^5×2^b/2^4=2^cA=B=C= Please I'm gonna fail math

Mathematics

1 answer:

aleksley [76]3 years ago

5 0

9514 1404 393

Answer:

a = 3, b = 12, c = 13

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c)

(a^b)^c = a^(bc)

___

You seem to have ...

$\dfrac{2^5\times8^4}{16}=\dfrac{2^5\times(2^3)^4}{2^4}\qquad (a=3)\\\\=\dfrac{2^5\times2^{3\cdot4}}{2^4}=\dfrac{2^5\times2^{12}}{2^4}\qquad (b=12)\\\\=2^{5+12-4}=2^{13}\qquad(c=13)$

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<em>Additional comment</em>

I find it easy to remember the rules of exponents by remembering that <em>an exponent signifies repeated multiplication</em>. It tells you how many times the base is a factor in the product.

$2\cdot2\cdot2 = 2^3\qquad\text{2 is a factor 3 times}$

Multiplication increases the number of times the base is a factor.

$(2\cdot2\cdot2)\times(2\cdot2)=(2\cdot2\cdot2\cdot2\cdot2)\\\\2^3\times2^2=2^{3+2}=2^5$

Similarly, division cancels factors from numerator and denominator, so decreases the number of times the base is a factor.

$\dfrac{(2\cdot2\cdot2)}{(2\cdot2)}=2\\\\\dfrac{2^3}{2^2}=2^{3-2}=2^1$

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Answer:

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Step-by-step explanation:

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3 years ago

What is the area of 1 times 6/7?

Rzqust [24]

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3 years ago

Read 2 more answers

Solve 10-5x+1=7x+11-12

Gekata [30.6K]

Answer:

x=1

Step-by-step explanation:

Let's solve your equation step-by-step.

10−5x+1=7x+11−12

Step 1: Simplify both sides of the equation.

10−5x+1=7x+11−12

10+−5x+1=7x+11+−12

(−5x)+(10+1)=(7x)+(11+−12)(Combine Like Terms)

−5x+11=7x+−1

−5x+11=7x−1

Step 2: Subtract 7x from both sides.

−5x+11−7x=7x−1−7x

−12x+11=−1

Step 3: Subtract 11 from both sides.

−12x+11−11=−1−11

−12x=−12

Step 4: Divide both sides by -12.

−12x

−12

x=1

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3 years ago

Plzzzzz help 3 9/10 is the same as the sum of 3 1/3 and half of a number what is the number

kirill [66]

Answer:

9/10 = 1/3 + 1/2x

9/10 - 1/3 = 1/2x

LCM of 10 and 3 is 30

( 27 - 10)/30 = 1/2x

17/30 = 1/2x

17 * 2 = 30x

34=30x

34/30= 30x/30

x = 17/15

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3 years ago

If X is a r.v. such that E(X^n)=n! Find the m.g.f. of X,Mx(t). Also find the ch.f. of X,and from this deduce the distribution of

astraxan [27]

$M_X(t)=\mathbb E(e^{Xt})$
$M_X(t)=\mathbb E\left(1+Xt+\dfrac{t^2}{2!}X^2+\dfrac{t^3}{3!}X^3+\cdots\right)$
$M_X(t)=\mathbb E(1)+t\mathbb E(X)+\dfrac{t^2}{2!}\mathbb E(X^2)+\dfrac{t^3}{3!}\mathbb E(X^3)+\cdots$
$M_X(t)=1+t+t^2+t^3+\cdots$
$M_X(t)=\displaystyle\sum_{k\ge0}t^k=\frac1{1-t}$

provided that $|t|$ .

Similarly,

$\varphi_X(t)=\mathbb E(e^{iXt})$
$\varphi_X(t)=1+it+(it)^2+(it)^3+\cdots$
$\varphi_X(t)=(1-t^2+t^4-t^6+\cdots)+it(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=(1+it)(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=\dfrac{1+it}{1+t^2}=\dfrac1{1-it}$

You can find the CDF/PDF using any of the various inversion formulas. One way would be to compute

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{e^{itx}\varphi_X(-t)-e^{-itx}\varphi_X(t)}{it}\,\mathrm dt$

The integral can be rewritten as

$\displaystyle\int_0^\infty\frac{2i\sin(tx)-2it\cos(tx)}{it(1+t^2)}\,\mathrm dt$

so that

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt$

There are lots of ways to compute this integral. For instance, you can take the Laplace transform with respect to $x$ , which gives

$\displaystyle\mathcal L_s\left\{\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt\right\}=\int_0^\infty\frac{1-s}{(1+t^2)(s^2+t^2)}\,\mathrm dt$
$=\displaystyle\frac{\pi(1-s)}{2s(1+s)}$

and taking the inverse transform returns

$F_X(x)=\dfrac12+\dfrac1\pi\left(\dfrac\pi2-\pi e^{-x}\right)=1-e^{-x}$

which describes an exponential distribution with parameter $\lambda=1$ .

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3 years ago