The graphs show the distances traveled by two bikers riding at constant rates.

tatyana61 [14]

Solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Step-by-step explanation:

We need to Solve the expression: $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

Solving:

$\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

$=\frac{4(3^3x^6y^{12})}{(2^4x^{12}y^{20})}\\=\frac{4(27x^6y^{12})}{(16x^{12}y^{20})}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}$

Applying exponent rule: $\frac{x^a}{x^b}=x^{a-b}$

$=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{27x^{-6}y^{-8}}{4}$

Another exponent rule says: $x^{-a}=\frac{1}{x^a}=$

$=\frac{27}{4x^{6}y^{8}}$

So, solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Keywords: Solving Exponents

Learn more about Solving Exponents at:

#learnwithBrainly

8 0

3 years ago

For f(x)= x+5 and g(x)=4x+2 <br> find (fog)(x) And (gof)(x)

Svetllana [295]

Composing functions means that the input of the outer functions is the output of the inner function.

In fact, you can rewrite the circle notation as

$(f\circ g)(x)=f(g(x)),\quad (g\circ f)(x)=g(f(x))$

So, we can substitute g(x) with its expression:

$(f\circ g)(x)=f(g(x))=f(4x+2)$

And since f(x)=x+5, we simply have to add 5 to its input:

$f(4x+2)=(4x+2)+5=4x+7$

Similarly, we have, substituting f with its expression,

$(g\circ f)(x)=g(f(x))=g(x+5)$

And since g(x)=4x+2, we have to multiply the input by 4 and add 2:

$g(x+5)=4(x+5)+2=4x+20+2=4x+22$

7 0

3 years ago

Bye guys im quitting, it was fun while it lasted thanks everybody for being here

nalin [4]

Answer:

Not sure if I was ever able to help you but I hope you have an amazing future<#

3 0

3 years ago

Suppose that the length of a phone call in minutes is an exponential random variable with parameter λ = 1/ 8. If someone arrives

Elis [28]

Answer:

a) P=0.535

b) P=0.204

c) P=0.286

Step-by-step explanation:

The exponential distribution is expressed as

$F(x>t)=e^{-\lambda t}$

In this example, λ=1/8=0.125 min⁻¹.

a) The probability of having to wait more than 5 minutes

$F(x>5)=e^{-0.125*5}=e^{-0.625}=0.535$

b) The probability of having to wait between 10 and 20 minutes

$F(1020)=e^{-0.125*10}-e^{-0.125*20}\\\\F(10$

c) The exponential distribution is memory-less, so it is independent of past events.

If you have waited 5 minutes, the probability of waiting more than 15 minutes in total is the same as the probability of waiting 15-5=10 minutes.

$F(x>15|t^*=5)=F(x>15)/F(x>5)=\frac{e^{-0.125*15}}{e^{-0.125*5}}=e^{-0.125*(15-5)}\\\\ F(x>15|t^*=5)=e^{-0.125*(10)}=F(x>10)=0.286$

5 0

3 years ago

Make x the subject please

Sauron [17]

Y^2 = (2x+1)/(x-1)
xy^2-y^2 = 2x + 1
xy^2 - 2x = y^2 + 1
x(y^2 - 2) = y^2 + 1
x = (y^2 + 1) / (y^2 - 2)

4 0

4 years ago

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