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2 years ago

.what is the factored form of 8x^24 - 27y^6

Mathematics

1 answer:

Thepotemich [5.8K]2 years ago

7 0

Answer:

I couldn't figure out how to type it so I typed it somewhere else lol. Hope it helps:)

Step-by-step explanation:

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Solve for q. -5(3-q) +4=5q-11

soldier1979 [14.2K]

Perform the indicated multiplication first: -15 + 5q + 4 = 5q - 11

Note that 5q appears on both sides of this equation. Cancelling, we get :

-11 = -11. This is always true. Thus, -5(3-q) +4=5q-11 is true for all q.

4 0

3 years ago

Write 25/176 as a decimal then tell if it's a terminating or a repeating decimal.

EastWind [94]

Answer:

0.14204545454545

It's a repeating decimal

Step-by-step explanation:

25/176 is the same as 25 ÷ 176

so 25 ÷ 176 = 0.14204545454545

and it is a repeating decimal because the reminder is not 0

8 0

3 years ago

A town with a population of 5,000 grows 3% per year. Find the population at the end of<br> 10 years.

TEA [102]

Answer:

6,500

Step-by-step explanation:

5000x.03=150

150=3%

150x10=1500

1,500+5,000= 6,500

5 0

3 years ago

What is -4 = s-6/-4 equaled to

zmey [24]

Move the negative sign to the left

-4 = s - (-6/4)

Simplify 6/4 to 3/2

-4 = s - (-3/2)

Simplify brackets

-4 = s + 3/2

Subtract 3/2 from both sides

-4 - 3/2 = s

Simplify -4 - 3/2 to -11/2

-11/2 = s

Switch sides

7 0

3 years ago

If a bacteria population starts with 80 bacteria and doubles every three hours, then the number of bacteria after t hours is n =

Thepotemich [5.8K]

Answer:

$f^{-1}(x)=3\frac{ln(\frac{t}{80})}{ln(2)}$

It will take 20.9 hours to reach 10000

Step-by-step explanation:

$f(t)=80(2)^{\frac{t}{3}}$

To find inverse of the function , replace f(x) with y

$y=80(2)^{\frac{t}{3}}$

swap the variables

$t=80(2)^{\frac{y}{3}}$

solve the equation for y

$t=80(2)^{\frac{y}{3}}\\\\\frac{t}{80} =(2)^{\frac{y}{3}}$

take ln on both sides

$ln(\frac{t}{80} )= \frac{y}{3} ln(2)\\\\\frac{ln(\frac{t}{80} }{ln(2)} =\frac{y}{3}\\3\frac{ln(\frac{t}{80} }{ln(2)} =y$

the inverse function is

$f^{-1}(x)=3\frac{ln(\frac{t}{80})}{ln(2)}$

for part b plug in 10000 for t in f^-1(x)

$f^{-1}(10000)=3\frac{ln(\frac{t}{80} }{ln(2)} \\f^{-1}(10000)=3\frac{ln(\frac{10000}{80} }{ln(2)} \\=20.89$

It will take 20.9 hours to reach 10000

7 0

4 years ago