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Assoli18 [71]
2 years ago
10

What are the domain, range, and asymptote of h(x) = (0.5)x – 9?

Mathematics
1 answer:
Rzqust [24]2 years ago
8 0

Let's solve for h

hx = 0.5x - 9

Step 1: Divide both sides by x.

hx/x = 0.5x - 9/ x

Answer = h = 0.5x - 9/ x

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What is the y value of the solution to the system of equations shown below y = 22 - 6 y = 50 – 21​
zhannawk [14.2K]

Answer:

22-6y= 29

-6y=29-22

-6y=7

y=-7/6

3 0
2 years ago
The median is the same as the ______________.
Ilia_Sergeevich [38]

Answer: Option A

The median is the same as the _ Second quartile_.

Step-by-step explanation:

Given a series of data ordered from least to greatest, the median is the value that is in the center.

That is, the median represents the value that divides 50% of the data.

In the same way, the first quartile Q_1 is the value that divides 25% of the data and the third quartile Q_3 is the value that divides 75% of the data.

The second quartile Q_2 is the number that divides 50% of the data.

Then notice that the second quartile is equal to the median

Then te answer is the option A.

7 0
3 years ago
Read 2 more answers
What is the domain and range ?
ElenaW [278]

Answer:

domain -4≤x≤4

range -1≤y≤2

Step-by-step explanation:

5 0
3 years ago
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A sample of a radioactive substance decayed to 97% of its original amount after a year. (Round your answers to two decimal place
Andrej [43]

Answer:

a) The half life of the substance is 22.76 years.

b) 5.34 years for the sample to decay to 85% of its original amount

Step-by-step explanation:

The amount of the radioactive substance after t years is modeled by the following equation:

P(t) = P(0)(1-r)^{t}

In which P(0) is the initial amount and r is the decay rate.

A sample of a radioactive substance decayed to 97% of its original amount after a year.

This means that:

P(1) = 0.97P(0)

Then

P(t) = P(0)(1-r)^{t}

0.97P(0) = P(0)(1-r)^{0}

1 - r = 0.97

So

P(t) = P(0)(0.97t)^{t}

(a) What is the half-life of the substance?

This is t for which P(t) = 0.5P(0). So

P(t) = P(0)(0.97t)^{t}

0.5P(0) = P(0)(0.97t)^{t}

(0.97)^{t} = 0.5

\log{(0.97)^{t}} = \log{0.5}

t\log{0.97} = \log{0.5}

t = \frac{\log{0.5}}{\log{0.97}}

t = 22.76

The half life of the substance is 22.76 years.

(b) How long would it take the sample to decay to 85% of its original amount?

This is t for which P(t) = 0.85P(0). So

P(t) = P(0)(0.97t)^{t}

0.85P(0) = P(0)(0.97t)^{t}

(0.97)^{t} = 0.85

\log{(0.97)^{t}} = \log{0.85}

t\log{0.97} = \log{0.85}

t = \frac{\log{0.85}}{\log{0.97}}

t = 5.34

5.34 years for the sample to decay to 85% of its original amount

8 0
3 years ago
Marco drove 75 miles in 1 2/3 hours. How many miles can he drive in 1 hour? (SHOW WORK)
love history [14]

Answer:

  45 miles

Step-by-step explanation:

Marco's average speed is ...

  (75 mi)/(5/3 h) = 75·3/5 mi/h = 45 mi/h

Marco's rate of travel is 45 miles in one hour.

4 0
3 years ago
Read 2 more answers
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