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3 years ago

Simplify ( g^3 a^4)^2

Mathematics

1 answer:

Alex777 [14]3 years ago

6 0

Just Distribute the squared to both g and a

$(g^3a^4)^2$

When Distributing an exponent outside the parenthesis you multiply the exponents

Which results in...

$g^6a^8$

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?/6 < 1/2 which one is it

Alekssandra [29.7K]

Answer:

1/6<1/2

Step-by-step explanation:

1/2 is .5 while 1/6 is .166 repeating

4 0

3 years ago

Read 2 more answers

36 out of 48 in percent form

aivan3 [116]

It would be .75 which equals 75%

8 0

3 years ago

Can someone please help me? I put the question below!

pogonyaev

Answer:

Step-by-step explanation:

the formula for the volume of a cone of base radius r and height h is:

V = (1/3)(πr²)(h). Here, r = 6 in and h = 30 in. Then,

V = (1/3)(π)(6 in)²(30 in), or

= (1/3)π(36 in²)(30 in), or

V = 360π in³

4 0

4 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

4 years ago

Find the area of the figure

yanalaym [24]

IT'S SImple just divid all them toghether 2x2x2x2x2x2x2x2x2=999,0000 and that is ur answer

8 0

3 years ago