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inn [45]
3 years ago
13

Determine if the equation y = 15(1.35)* represents exponential growth or decay and justify. Hint: Think about the growth and dec

ay equations.​
Mathematics
2 answers:
Sever21 [200]3 years ago
6 0
<h3>Answer: Exponential growth</h3>

====================================================

Explanation:

The equation is in the form y = a*(b)^x

The 'a' is the initial or starting value. In this case, a = 15.

The b is the base of the exponent which is tied to growth or decay. If 0 < b < 1, then we have decay. If b > 1, then we have growth.

Since b = 1.35, and 1.35 > 1, this means b > 1 and the given equation represents exponential growth.

Solving 1+r = 1.35 leads to r = 0.35. This means the growth rate is 35%

siniylev [52]3 years ago
3 0

Answer:sdfghgfdfghnhgfdfgh

Step-by-step explanatsdfghjhgfghjk

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15.3+x=1.3-x so how do I solve it
anzhelika [568]

Answer:

-7

Step-by-step explanation:

Let's try getting rid of the numbers to isolate the x.

Add x on both sides to get rid of the x on the right side and to get 2x on the left side. This leaves us with 15.3 + 2x = 1.3

Subtract 15.3 on both sides to get rid of the 15.3 on the left side.

This would leave us with 2x = -14

Divide 2 on both sides to isolate the x.

The answer is x = -7

3 0
4 years ago
Read 2 more answers
Please show work and be done soon
stich3 [128]

9514 1404 393

Answer:

  x = 22.5°

Step-by-step explanation:

Angle BED ≅ angle EBC = 6x, so angle BEA = 4x. The angles FEB and BEA form a linear pair, so we have ...

  4x +4x = 180°

  x = 22.5° . . . . . . divide by 8

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BED and EBC are "alternate interior angles" with respect to transversal BE crossing parallel lines DE and AC.

8 0
3 years ago
At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is
Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = \frac{dx}{dt}

ship B is sailing north at 20 km/h =\frac{dy}{dt}

here x and y are the  sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

                 y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find \frac{dz}{dt} at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we |  | get \\

z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h

derivative first equation w . r. to t we get

2z\frac{dz}{dt} =-2(130-x)\frac{dx}{dt}+2y\frac{dy}{dt}

\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]

\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }

     = \frac{1100}{85.44}\\  = 12.87km/h

8 0
3 years ago
Can anyone help me please
zaharov [31]

Answer:

1. is 45/2

2. is 12/5

3. is 4/5

4. is 14

5. is 28

6. is 6         hope this helps :)

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Solve: -2(4g - 3) = 30 *<br> 1 point<br> g = 4<br> g = -4<br> g = 2<br> g = -3
Paraphin [41]

-2(4g - 3) = 30\\-8g+6=30\\-8g=24\\g=-3

P.S. Hello from Russia

8 0
3 years ago
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