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Klio2033 [76]
2 years ago
15

Find the domain and range of f(x)=4*(1/2)^x

Mathematics
1 answer:
jolli1 [7]2 years ago
6 0
Domain is the set of x-values and range is the set of y-values.

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What is the prime factorization for 900 i just need the answer i got to turn in in like 2 minute ​
Tomtit [17]

Answer:

Factors (2 x 2 x 3 x 3 x 5 x 5 = 900). It can also be written in exponential form as 22 x 32 x 52.

Step-by-step explanation:

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Match the description with the key term it represents. 1. Binder containing transcripts, diplomas, recommendations, etc. portfol
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Answer:

1. portfolio

2. résumé

3. online portfolio

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I need help pleaseeeeeee
enyata [817]

Answer:

Its B and D

Step-by-step explanation:

If you have any questions use Math-way it realy helps

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3 years ago
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The points in the table lie on a line. Find the slope of the line.
Wewaii [24]

Answer:

Slope (m) = -\frac{3}{4}

Step-by-step explanation:

Using any two pairs from the table of values given, (-2, 5) and (2, 2):

Slope (m) = \frac{y_2 - y_1}{x_2 - x_1}

Where,

(-2, 5) = (x_1, y_1)

(2, 2) = (x_2, y_2)

Plug in the value into the slope formula:

Slope (m) = \frac{2 - 5}{2 -(-2)}

Slope (m) = -\frac{3}{4}

8 0
3 years ago
A basin is filled by two pipes in 12 minutes and 16 minutes respectively. Due to the obstruction of water flow after the two pip
olasank [31]

Answer:

The time duration of the two pipes restricted flow before the flow became normal is 4.5 minutes

Step-by-step explanation:

The given information are;

The time duration for the volume, V, of the basin to be filled by one of the pipe, A, = 12 minutes

The time duration for the volume, V, of the basin to be filled by the other pipe, B, = 16 minutes

Therefore, the flow rate of pipe A = V/12

The flow rate of pipe B = V/16

Due to the restriction, we have;

The proportion of its carrying capacity the first pipe, A, carries = 7/8 of the carrying capacity

The proportion of its carrying capacity the second pipe, B, carries = 5/6 of the carrying capacity

Whereby the tank is filled 3 minutes after the restriction is removed, we have;

\dfrac{7}{8} \times \dfrac{V}{12} \times t + \dfrac{5}{6} \times \dfrac{V}{16} \times t +  \dfrac{V}{12} \times 3 + \dfrac{V}{16} \times 3 = V

Simplifying gives;

\dfrac{(2\cdot t +7) \cdot V}{16}  = V

2·t + 7 = 16

t = (16 - 7)/2 = 4.5 minutes

Therefore, it took 4.5 seconds of the restricted flow before the the flow of water in the two pipes became normal

7 0
3 years ago
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