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3 years ago

Which function below is decaying the fastest?

Mathematics

1 answer:

a_sh-v [17]3 years ago

7 0

Answer:

Function: y = 1/6^x

Step-by-step explanation:

The function y = 1/6^x would be decaying faster because the function y = 2/3^x equals 4/6^x and that is 4 times greater than 1/6. That means that more of the value will be retained in the function y = 2/3^x or y = 4/6^x.

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A farm produces an average of 17.1 liters of milk per day. About how many days will it take to produce 200 liters of milk? A. ab

Aleks04 [339]

Average produce milk per day 17.1 liters
to produce 200 liters should be:

17.1x = 200
x = 200/17.1
x = 11,69

it close to 12. the answer is a. about 12 days

hope this help

6 0

3 years ago

Read 2 more answers

Polar coordinates of a point are given. Find the rectangular coordinates of the point. (2.1 ; 2pi/9)

maxonik [38]

It is b. your answer is B

4 0

3 years ago

Look at picture (15 point)

NNADVOKAT [17]

Answer:

Option C is the correct option

Step by step explanation

$\sqrt{ \frac{22 {x}^{6} }{11 {x}^{4} } }$

Reduce the fraction with 11

$\sqrt{ \frac{2 {x}^{6} }{ {x}^{4} } }$

Simplify the expression

$\sqrt{2 {x}^{6 - 4} }$

$\sqrt{2 \: {x}^{2} }$

Simplify the radical expression

$x \sqrt{2}$

Hope this helps...

Good luck on your assignment..

3 0

2 years ago

Pleassssssseeeeeeeeeeeeeese

Finger [1]

The ratio is 12:28. Or in fraction form it’s 12/28. You can also simplify it to 3/7

4 0

3 years ago

Read 2 more answers

a toy rocket is launched from the top of a 48 foot hill. The rockets initial upward velocity is 32 feet per second and its heigh

sdas [7]

Answer:

t = 2.11 seconds

Step-by-step explanation:

A toy rocket is launched from the top of a 48 foot hill. The rockets initial upward velocity is 32 feet per second and its height h at any given second t is modeled by the equation:

$h=-16t^2+32t+48$

Let us assume that we need to find the time by it to reach the ground. It means h = 0

$-16t^2+32t+48=0$

The above is a quadratic equation. The value of t is given by :

$t=\dfrac{-b\pm \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-b+ \sqrt{b^2-2ac} }{2a},\dfrac{-b- \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-32+ \sqrt{(32)^2-2\times (-16)(8)} }{2\times (-16)},\dfrac{-32-\sqrt{(32)^{2}-2\times(-16)(8)}}{2\times(-16)}\\\\t=-0.11\ s, 2.11\ s$

So, it will take 2.11 seconds to reach the ground.

6 0

3 years ago