Answer:
only the first option is equivalent
Step-by-step explanation:
Usual limit of sin is sinX/X--->1, when X--->0
sin3x/5x^3-4x=0/0?, sin3x/3x--->1 when x --->0, so sin3x/5x^3-4x= [3x. sin3x / 3x] /(5x^3-4x)=(sin3x / 3x) . (3x/5x^3-4x)
=(sin3x / 3x) . (3/5x^2- 4)
finally lim sin3x/5x^3-4x=lim (sin3x / 3x) .(3/5x^2- 4)=1x(3/-4)= - 3/4
x----->0 x---->0
Answer:
h = -14
Step-by-step explanation:
-7πh = 98π
⇔ -7h = 98 (Just Divide both sides by π)
⇔ h = 98 ÷ (-7) = -14
Answer:
<h2>The area of the base is 144 square inches.</h2><h2>The area of each triangular face is 66 square inches.</h2><h2>Grabiel needs 408 square inches of paint.</h2>
Step-by-step explanation:
The complete problem is attached.
Notice that the figure is a square pyramid, where its base dimensions are 12 inches by 12 inches, which represents an area of

The slant height of the pyramid is 11 inches, which allow us to find the area of each triangle face

But there are four triangle faces, so
.
Therefore, the area of each triangular face is 66 square inches.
So, the total surface area would be the sum

Therefore, Gabriel needs 408 square inches to paint the whole model.
Answer:
The home would be worth $249000 during the year of 2012.
Step-by-step explanation:
The price of the home in t years after 2004 can be modeled by the following equation:

In which P(0) is the price of the house in 2004 and r is the growth rate.
Since 2003 median home prices in Midvale, UT have been growing exponentially at roughly 4.7 % per year.
This means that 
$172000 in 2004
This means that 
What year would the home be worth $ 249000 ?
t years after 2004.
t is found when P(t) = 249000. So







2004 + 8.05 = 2012
The home would be worth $249000 during the year of 2012.