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Vedmedyk [2.9K]
3 years ago
7

What’s the solution for this problem???

Mathematics
1 answer:
Gre4nikov [31]3 years ago
4 0
The correct answer is C.
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mina [271]

it is d. Pensacola beach

3 0
3 years ago
Each day 1/2 of the money that is in bank value is removed. No money is added to the value. Which of the following represent the
Alchen [17]

Answer:

Option C. exponential decay function

Step-by-step explanation:

In this problem we have a exponential function of the form

y=a(b^x)

where

y ---> is the money that is in the bank

x ----> number of days

a ---> is the initial value or y-intercept

b is the base of the exponential function

r ---> is the rate of change

b=(1+r)

In this problem we have that

Each day 1/2 of the money that is in bank value is removed

so

r=-50\%=-50/100=-0.50 ---> is negative because is a decreasing function

The value of b is equal to

b=1-r=1-0.50=0.50

y=a(0.50^x)

The value of b is less than 1

b<1

That means -----> Is a exponential  decay function

3 0
3 years ago
Evaluate log base 12 of y^2, given log base 12=16
Ostrovityanka [42]

Answer:

y = 12^8

Step-by-step explanation:

Log base 12, y^2 = 16

y2 = 12^16

y = sqrt ( 12^16) = (12^16)^1/2

y = 12^8

5 0
3 years ago
The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.
VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

\int\limits^2_{-1} {(f(x) - g(x))} \, dx

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx

We know that:

\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}

\int\limits^{}_{} {1} \, dx = x

\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}

Then our integral is:

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2}  + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3}  + (-1) - \frac{(-1)^2}{2}  )

The right side is equal to:

(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6

The bounded area is 5 + 5/6 square units.

3 0
3 years ago
Say you have a tic toe box of 3 by 3 how do you make each diagonal and vertical line equal 15 only using # 1-10
SpyIntel [72]

| 8 |  | 1 |  | 6 |
-------------------
| 3 |  | 5 |  | 7 |
-------------------
| 4 |  | 9 |  | 2 |

7 0
4 years ago
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