4000 Is blank times as 4

Match each exponential function to its percent rate of change.

vredina [299]

This question is about exponent function. All the function in this question is following a pattern of f(x)= a $(b)^{x}$ In this function, a is the initial/starting quantity and b is the base of the exponent. The option in this problem is about growth/decay that was determined by the base of the exponent. So, to answer this question you just need to pay attention to the variable b

1. Answer: 4% grow
f(x)= a $(b)^{x}$
f(x)= 46(1.04)²
Then the value of the variable would be:
a= 46
b=1.04
Since b is >1 then it is a growing function. The grow in percent would be: (1.04 * 100%) - 100%= 104%-100%=4%

2. Answer: 4% decay
f(x)= a $(b)^{x}$
f(x)= 104(0.96)²
Then the value of the variable would be:
a= 104
b=0.96
Since b is <1 then the function would decay. The rate of change percent would be: (.96 * 100%) - 100%= 96%-100%= -4%. The function rate of change is 4% decay

3. Answer: 40% decay
f(x)= a $(b)^{x}$
f(x)= 74(0.6)²
Then the value of the variable would be:
a= 74
b=0.60
Since b is <1 then the function would decay. The rate of change percent would be: (0.60 * 100%) - 100%= 60%-100%= -40%. The function rate of change is 40% decay

4. Answer: growth 40%
f(x)= a $(b)^{x}$
f(x)= 44(1.4)²
Then the value of the variable would be:
a= 44
b=1.4
Since b is >1 then the function would grow. The rate of change percent would be: (1.40 * 100%) - 100%= 140%-100%= 40%. The function rate of change is 40% growth

5. Answer: 14% decay
f(x)= a $(b)^{x}$
f(x)= 40(0.86)²
Then the value of the variable would be:
a= 40
b=0.86
Since b is <1 then the function would decay. The rate of change percent would be: (0.86 * 100%) - 100%= 86%-100%= -14%. The function rate of change is 14% decay

6. Answer: 14% growth
f(x)= a $(b)^{x}$
f(x)= 8(1.14)²
Then the value of the variable would be:
a= 8
b=1.14
Since b is >1 then the function would grow. The rate of change percent would be: (1.14 * 100%) - 100%= 114%-100%= 14%. The function rate of change is 14% growth

3 0

3 years ago

WILL GIVE A BRAINLEST<br><br> Which piecewise relation defines a function?

HACTEHA [7]

The 3rd Image defines a piecewise function because for it to be a function, every input must match to exactly one and only one output. In Images 1, 2, and 4, there are certain inputs that have two outputs or stated otherwise, have two y-values for the same x-value. Only the 3rd Image matches 1 x-value to every 1 y-value. So, that's your answer.

4 0

3 years ago

Read 2 more answers

Parabola passes through the points (0,5), (1,4), and (2,5). What function does the graph represent?

AfilCa [17]

Answer:

x^2 - 2x + 5 = 0.

Step-by-step explanation:

The (0, 5) is the point where the parabola passes through the y axis (where x = 0), so we can write the equation as

y = ax^2 + bx + 5 where a and b are constants to be found.

Also, since (1, 4) and (2, 5) are points on the curve, substituting, we have the system:

a(1)^2 + 1b + 5 = 4

a(2)^2 + 2b + 5 = 5

Simplify these 2 equations:

a + b = -1 .................(1)

4a + 2b = 0..................(2)

Multiply the first equation by -2:

-2a - 2b = 2 .................(3)

Add (2) + (3):

2a = 2

a = 1.

Substitute a = 1 into (2):-

4*1 + 2b = 0

2b = -4

b = -2.

7 0

3 years ago

How can I explain with mental math

jeka94

Use a property that’s a good reason then tear it apart

6 0

3 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

The positive value of $k$ will result in exactly one real root is approximately 0.028.

7 0

2 years ago