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nadezda [96]
3 years ago
5

In exponential growth functions, the base of the exponent must be greater than 1. How would the function change if the base of t

he exponent were 1? How would the function change if the base of the exponent were between 0 and 1?
Sample Response: If the base of the exponent were 1, the function would remain constant. The graph would be a horizontal line. If the base of the exponent were less than 1, but greater than 0, the function would be decreasing.

What did you include in your response? Check all that apply.

If the base were 1, the function would be constant.
If the base were 1, the graph would be a horizontal line.
If the base were between 0 and 1, the function would be decreasing.
Mathematics
1 answer:
alexandr402 [8]3 years ago
7 0

Answer:

How would the function change if the base of the exponent were between 0 and 1? If the base of the exponent were 1, the function would remain constant. The graph would be a horizontal line. If the base of the exponent were less than 1, but greater than 0, the function would be decreasing.

Step-by-step explanation:

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dimulka [17.4K]

Answer:

12

Step-by-step explanation:

side 1: 4

side 2: 6

6*4*(1/2)

12

6 0
3 years ago
Solve for Z.<br> 3z - 12 =48
marysya [2.9K]

Answer:

Z = 20

Step-by-step explanation:

3 x 20 = 60

60 - 12 = 48

4 0
2 years ago
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Which line of symmetry for the parabola (x-1)^2= 4(y-1)^?
jonny [76]
This is a vertical parabola, because  (x-1)².
Vertex of the parabola (1,1).
So line symmetry is x=1.
4 0
2 years ago
A ship moves through the water at 30 miles/hour at an angle of 30° south of east. The water is moving 5 miles/hour at an angle o
iVinArrow [24]

Answer:

a. 25.98i - 15j mi/h

b. 1.71i + 4.7j mi/h

c. 27.69i -10.3j mi/h

Step-by-step explanation:

a. Identify the ship's vector

Since the ship moves through water at 30 miles per hour at an angle of 30° south of east, which is in the fourth quadrant. So, the x-component of the ship's velocity is v₁ = 30cos30° = 25.98 mi/h and the y-component of the ship's velocity is v₂ = -30sin30° = -15 mi/h

Thus the ship's velocity vector as ship moves through the water v = v₁i + v₂j = 25.98i + (-15)j = 25.98i - 15j mi/h

b. Identify the water current's vector

Also, since the water is moving at 5 miles per hour at an angle of 20° south of east, this implies that it is moving at an angle 90° - 20° = 70° east of north, which is in the first quadrant. So, the x-component of the water's velocity is v₃ = 5cos70° = 1.71 mi/h and the y-component of the water's velocity is v₄ = 5sin70° = 4.7 mi/h

Thus the ship's velocity vector v' = v₃i + v₄j = 1.71i + 4.7j mi/h

c. Identify the vector representing the ship's actual motion.

The velocity vector representing the ship's actual motion is V = velocity vector of ship as ship moves through water + velocity vector of water current.

V = v + v'

= 25.98i - 15j mi/h + 1.71i + 4.7j mi/h

= (25.98i + 1.71i + 4.7j - 15j )mi/h

= 27.68i -10.3j mi/h

7 0
2 years ago
Tyler delivers newspapers.He collected $10 from each customer.He has 5 customers on each street,and delivers to 4 different stre
strojnjashka [21]

Answer:

$200

Step-by-step explanation:

4 different streets with 5 customers on each means 4*5 customers or 20 customers. If he has 20 customers and he gets $10 from each one, thats 20*10. he collected $200

6 0
2 years ago
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