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Stels [109]
4 years ago
12

Which function below is decaying the fastest?

Mathematics
1 answer:
a_sh-v [17]4 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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If sinθ = 12/13 and θ is an acute angle, find cotθ.
Tju [1.3M]

Answer:

cot(θ)=5/12

Step-by-step explanation:

(I'm going to replace theta with x to save time.)

Recall that sine is equal to opposite over hypotenuse. Thus:

sin(x)=12/13, this means that the opposite side is 12 while the hypotenuse is 13 (well, it doesn't have to be, but their ratios are).

Find the adjacent side using the Pythagorean Theorem.

13^2 = 12^2 + adj^2

adj^2 = 25

adj = 5.

Recall that tan(x) is opposite over adjacent. This means that cot(x) is adjacent is opposite.

Therefore: cot(x) = 5/12.

6 0
4 years ago
A little girl kicks a soccer ball. It goes 10 feet and comes back to her. How is this possible?
Step2247 [10]
The little girl kicked the call in the air and it came back down
4 0
3 years ago
Read 2 more answers
Find the product. -8x5y2 · 6x2y
katrin [286]

Answer:

The product of the given expression -8x^5y^2\times 6x^ 2y= 72x^6y^3[/tex]

Step-by-step explanation:

Given expression is -8x^5y^2\times 6x^2y

To find the product of the given expression:

The product of the given expression can be written as

-8x^5 y^2\times 6x 2y=(-8x^5y^2)(6x^2y)

=(-8x^5y^2)(6x^2y)

=-48(x^5\times x^2)(y^2\times y^1)  [since a^m\times a^n=a^{m+n}]

=-48x^7y^3

Therefore the product of the given expression -8x^5y^2\times 6x^2y is -48x^7y^3

Therefore,  -8x^5y^2\times 6x^2y=-48x^7y^3[/tex]

3 0
3 years ago
The area of the Pacific Ocean is more than 60 million square miles.
garri49 [273]

The scientific notation for this is :-6x10⁷

5 0
3 years ago
The range of the function f(k) = k2 + 2k + 1 is {25, 64}. What is the function’s domain?
s344n2d4d5 [400]
{k^2 + 2k + 1 = 25, k^2 + 2k + 1 = 64}
{k^2 + 2k - 24 = 0, k^2 + 2k - 63 = 0}
{(k - 4)(k + 6) = 0, (k - 7)(k + 9) = 0}
{k = -6, 4, -9, 7)
Therefore, range = {-9, -6, 4, 7)
3 0
4 years ago
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