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2 years ago

What can you say about the end behavior of the function f(x)=log10(5x-1)

2 answers:

6 0

The given function is
f(x) = log₁₀(5x-1)

As x -> -∞, the argument of the log function becomes a large negative number.
Because the log of a negative number is undefined, f(x) is undefined as x -> -∞.

As x -> +∞, the argument of the log function becomes a large positive number.
Therefore f(x) -> +∞ as x -> +∞.

Answer:
As x -> -∞, f(x) is undefined.
As x-> +∞, f(x) -> +∞.

Rainbow [258]2 years ago

6 0

Answer:

one end increases and one end decreases (Apex answer)

Goodluck everyone!!! :)

Step-by-step explanation:

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On average, water flows over a particular water fall at a rate of 2.05 x 105 cubic feet per second. One cubic foot of water weig

iren [92.7K]

Answer:

5.024 × 10^8 tons /day

Step-by-step explanation:

This question has to deal with conversion

We are told in the question that:

On average, water flows over a particular water fall at a rate of 2.05 x 10^5 cubic feet per second.

Water flow rate = 2.05 × 10^5 ft³/s

From the question,

One cubic foot of water weighs 62.4 lb.

1 ft³ = 62.4 lb

1 ton = 2200 lb

We are to calculate the flow rate in tons per day

Hence,

1 ft³ = 62.4 lb

2.05 × 10^5 ft³ =

Cross Multiply

2.05 × 10^5 × 62.4

= 12792000lb

Hence the water flow rate = 12792000lb/s

From the question,

2200 lb = 1 ton

12792000 Ib =

Cross Multiply

12792000lb × 1 ton/2200 lb

= 5814.5454545 tons

Hence the water flow rate = 5814.5454545 tons/second

We want the flow rate to be in tons/day

1 ton / second = 86400 tons / day

5814.5454545 tons/second = x tons/ day

Cross Multiply

x = 5814.5454545 tons/second × 86400 tons /day/ 1 ton / second

= 502376727.27 tons/day

= 5.0237672727 × 10^8 tons/day

Approximately ≈ 5.024 × 10^8 tons /day

Therefore, the rate of water flow in tons of water per day is 5.024 × 10^8 tons of water /day

7 0

2 years ago

Two boats leave port at noon. Boat 1 sails due east at 12 knots. Boat 2 sails due south at 8 knots. At 2 pm the wind diminishes

Ivan

Answer:

14.86 knots.

Step-by-step explanation:

<em>Given that:</em>

The boats leave the port at noon.

Speed of boat 1 = 12 knots due east

Speed of boat 2 = 8 knots due south

At 2 pm:

Distance traveled by boat 1 = 24 units due east

Distance traveled by boat 2 = 16 units due south

Now, speed of boat 1 changes to 9 knots:

At 3 pm:

Distance traveled by boat 1 = 24 + 9= 33 units due east

Distance traveled by boat 2 = 16+8 = 24 units due south

Now, speed of boat 1 changes to 8+7 = 15 knots

At 5 pm:

Distance traveled by boat 1 = 33 + 2 $\times$ 9= 51 units due east

Distance traveled by boat 2 = 24 + 2 $\times$ 15 = 54 units due south

Now, the situation of distance traveled can be seen by the attached right angled $\triangle AOB$ .

O is the port and A is the location of boat 1

B is the location of boat 2.

Using pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AB^{2} = OA^{2} + OB^{2}\\\Rightarrow AB^{2} = 51^{2} + 54^{2}\\\Rightarrow AB^{2} = 2601+ 2916 = 5517\\\Rightarrow AB = 74.28\ units$