Which equation represents exponential decay?

Mathematics

1 answer:

MAVERICK [17]3 years ago

6 0

Y=(1/5)^x would be the only equation demonstrating decay of these

You might be interested in

Need help with multiple answer math problem!

svetlana [45]

(a) getting the difference of outputs of any two inputs that are 1 unit apart, we have the following:
-3 – (-11) = 8
5 – (-3) = 8
13 – 5 = 8
21 – 13 = 8
29 – 21 = 8
37 – 29 = 8
45 – 37 = 8
53 – 45 = 8
(b) getting the difference of outputs of any two inputs that are 2 units apart, we have the following:
5 – (-11) = 16
13 – (-3) = 16
21 – 5 = 16
29 – 13 = 16
37 – 21 = 16
45 – 29 = 16
53 – 37 = 16
(c) getting the difference of outputs of any two inputs that are 3 units apart, we have the following:
13 – (-11) = 24
21 – (-3) = 24
29 – 5 = 24
37 – 13 = 24
45 – 21 = 24
53 – 29 = 24
(d) To get the ratio of the required problem, first we divide the first difference of outputs by one input interval, that is
8÷1=8
Next, we divide the second differences of outputs by two input intervals, that is
16÷2=8
Lastly, we divide the third differences of outputs by three input intervals, that is
24÷3=8
Notice that each ratio have the same result which is 8. This means that whatever the number input intervals are, between two outputs is 8 times apart.

6 0

3 years ago

What is the distance between -a and 0

Mashutka [201]

Answer:

The answer is: a

Step-by-step explanation:

0 - a = -a. So -a + a = 0

4 0

3 years ago

What is the image of (1, -6) for a 270° counterclockwise rotation about the origin?

gulaghasi [49]

I think (-1,-6)
hope this help

7 0

3 years ago

Read 2 more answers

Find symmetric equations for the line of intersection of the planes. z = 4x − y − 13, z = 6x + 5y − 13

MArishka [77]

The intersection line of two planes is the cross product of the normal vectors of the two planes.

p1: z=4x-y-13 => 4x-y-z=13
p2: z=6x+5y-13 => 6x+5y-z=13
The corresponding normal vectors are:
n1=<4,-1,-1>
n2=<6,5,-1>

The direction vector of the intersection line is the cross product of the two normals,
vl=
i j k
4 -1 -1
6 5 -1
=<1+5, -6+4, 20+6>
=<6,-2,26>
We simplify the vector by reducing its length by half, i.e.
vl=<3,-1,13>

To find the equation of the line, we need to find a point on the intersection line.
Equate z: 4x-y-13=6x+5y-13 => 2x+6y=0 => x+3y=0.
If x=0, then y=0, z=-13 => line passes through (0,0,-13)

Proceed to find the equation of the line:
L: (0,0,-13)+t(3,-1,13)
Convert to symmetric form:
$\frac{x-0}{3}=\frac{y-0}{-1}=\frac{z-(-13)}{13}$
=>
$\frac{x}{3}=\frac{-y}{1}=\frac{z+13}{13}$

3 0

3 years ago

Please helppppppppp me

Iteru [2.4K]

So there was a total of 28 games played and the Houston Tigers only scored above 53 points 4 times. Now the fraction would be 4/28 but we have to reduce so the real answer is actually 1/7

5 0

3 years ago