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luda_lava [24]
1 year ago
13

Given the following exponential function, identify whether the change represents

Mathematics
1 answer:
GarryVolchara [31]1 year ago
3 0

The exponential equation represents an exponential growth because the rate of decay is 1.02 which is greater than 1.

y = 90(1.02)^x

The general form equation is:

y(x)= a(1+r)^x such that r is the decay percent.

on comparing,

1 + r = 1.02

r = 1.02 - 1

r = 0.02

converting into percent

r = 0.02 * 100

r = 2%.

Hence the exponential growth is 2%.

Learn more about the exponential growth here:

brainly.com/question/11487261

#SPJ1

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How many solutions does this equation have?
Leto [7]

Answer:

The given equation having only one solution i.e., "w = 3". A further explanation is provided below.

Step-by-step explanation:

The given equation is:

⇒ 9(w-9)-2=-7w+7(w-8)

By opening the brackets, we get

⇒ 9w-81-2=-7w+7w-56

⇒ 9w-83=-56

On adding "83" both sides, we get

⇒ 9w-83+83=-56+83

⇒                 9w=27

⇒                   w=\frac{27}{9}

⇒                       =3

4 0
3 years ago
Evaluate 5^28·5^-32·5^-4<br> A. 17,920<br> B. 0.017920<br> C. 0.000256<br> D. 0.00000256
romanna [79]

Answer:

D 0.00000256

Step-by-step explanation:

5^{28} *5^{-32}*5^{-4}=5^{28-32-4}=5^{-8}=\frac{1}{5^{8}} =0.00000256

4 0
3 years ago
Read 2 more answers
The radius of the circle is increasing at a rate of 2 meters per minute and the sides of the square are increasing at a rate of
Lunna [17]

Answer:

Change in area=24\pi-48

Step-by-step explanation:

Let s will be the side of square and r will be the radius of circle.

Then two given conditions are

1)dr/dt=2 m/s

2)ds/dt=1 m/s

Area enclosed=(Area of square)-(Area of circle)

Area of square=s^{2}

Area of circle=\pi r^{2}

Area enclosed=(\pi  r^{2})-s^{2}

dA/dt=2\pir(dr/dt)-2s(ds/dt)

At s=24,and r=6

dA/dt=2(\pi)(6)(2)-2(24)(1)

Change in area=24\pi-48

6 0
3 years ago
PLLLZ HELP I AM GIVING 50 POINTS IF U ANSWER
Elden [556K]
A. the lower extreme is an outer but the upper exterme is not
6 0
3 years ago
Read 2 more answers
A city planner designs a park that is a quadrilateral with vertices at J(-3,1), K(1,3), L(5,-1), and M(-1,-3). There is an entra
Ksivusya [100]
The Quadrilateral is JKLM, 

let M_{JK}, M_{KL}, M_{LM}, M_{JM}, be the midpoints of JK, KL, LM and JM respectively.

---------------------------------------------------------------------------------------------------------

Given any 2 point P(m,n) and Q(k,l),<span>

the coordinates of the midpoint of the line segment PQ are given by the formula:

M_{PQ}=( \frac{m+k}{2} ,&#10;\frac{n+l}{2}), </span>

-------------------------------------------------------------------------------------------------

thus the coordinates of points M_{JK}, M_{KL}, M_{LM}, M_{JM},

are as follows:

M_{JK}= (\frac{-3+1}{2}, \frac{1+3}{2})=(-1,2), \\\\M_{KL}= (\frac{1+5}{2}, \frac{3-1}{2}=(3,1), \\\\M_{LM}= (\frac{5-1}{2}, \frac{-1-3}{2})=(2, -2),\\\\ M_{JM}= (\frac{-3-1}{2}, \frac{1-3}{2})=(-2,-1)


------------------------------------------------------------------------------------------------

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

 |PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}&#10;}</span>

------------------------------------------------------------------------------------------------

thus the distances connecting the opposite entrances can be calculated as follows:


|M_{JK},M_{LM}|= \sqrt{ (-1-2)^{2} + (2-(-2))^{2} }= \sqrt{9+16}=5

|M_{KL}M_{JM}|= \sqrt{ (3-(-2))^{2} + (1-(-1))^{2}}= \sqrt{25+4}= \sqrt{29}=5.39


Thus the total distance of the paths joining the opposite entrances is 

5+5.39 units = 50 m + 53.9 m = 104 m (rounded to the nearest meter)


Answer: 104 m



8 0
3 years ago
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