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slega [8]
2 years ago
10

Describe how to transform the graph of g(x)=logx into the graph of f(x)=2log(4-x).

Mathematics
1 answer:
elena-s [515]2 years ago
5 0

Answer:

I'm not sure if you have answer choices but here's the explaination!!

Step-by-step explanation:

Reflect across the y-axis, translate 4 units to the right, and vertically stretch by a factor of 2.  

Hope this helps!! Good luck. (:

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Is 10/3 irrational or rational
Masteriza [31]

It's rational


Rational numbers are numbers that you can put into the form of a/b.

7 0
3 years ago
3*5-4<br> ------------------------------
exis [7]
3*5=15 
15-4=11
therefore 11 is your answer
8 0
3 years ago
Read 2 more answers
A culture started with 4,000 bacteria. After 5 hours, it grew to 4,800 bacteria. Predict how many bacteria will be present after
eimsori [14]
First we need to find k ( rate of growth)
The formula is
A=p e^kt
A future bacteria 4800
P current bacteria 4000
E constant
K rate of growth?
T time 5 hours
Plug in the formula
4800=4000 e^5k
Solve for k
4800/4000=e^5k
Take the log for both sides
Log (4800/4000)=5k×log (e)
5k=log (4800/4000)÷log (e)
K=(log(4,800÷4,000)÷log(e))÷5
k=0.03646

Now use the formula again to find how bacteria will be present after 15 Hours
A=p e^kt
A ?
P 4000
K 0.03646
E constant
T 15 hours
Plug in the formula
A=4,000×e^(0.03646×15)
A=6,911.55 round your answer to get 6912 bacteria will be present after 15 Hours

Hope it helps!
4 0
3 years ago
Read 2 more answers
2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One
Stells [14]

y'=(t+y)^2-1

Substitute u=t+y, so that u'=y', and

u'=u^2-1

which is separable as

\dfrac{u'}{u^2-1}=1

Integrate both sides with respect to t. For the integral on the left, first split into partial fractions:

\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1

\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt

\dfrac12(\ln|u-1|-\ln|u+1|)=t+C

Solve for u:

\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C

\ln\left|1-\dfrac2{u+1}\right|=2t+C

1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}

\dfrac2{u+1}=1-Ce^{2t}

\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}

u=\dfrac2{1-Ce^{2t}}-1

Replace u and solve for y:

t+y=\dfrac2{1-Ce^{2t}}-1

y=\dfrac2{1-Ce^{2t}}-1-t

Now use the given initial condition to solve for C:

y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}

so that the particular solution is

y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}

3 0
3 years ago
Solve the equation p^2 + 4p = 1 by completing the square.
igor_vitrenko [27]

Answer:

p = -2 ±sqrt( 5)

Step-by-step explanation:

p^2 + 4p = 1

Take the coefficient of p

4

Divide by 2

4/2 =2

Square it

2^2 = 4

Add it to each side

p^2 + 4p+4 = 1+4

(p+2) ^2 = 5

Take the square root of each side

sqrt((p+2) ^2) =±sqrt( 5)

p+2 = ±sqrt( 5)

Subtract 2 from each side

p+2-2 = -2 ±sqrt( 5)

p = -2 ±sqrt( 5)

7 0
2 years ago
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