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3 years ago

Given the expression 12C7, which of the following scenarios could this combination represent?

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

6 0

Answer: B

<u>Step-by-step explanation:</u>

₁₂C₇ is read as "12 choose 7" which means you are choosing 7 out of 12. Since it is a combination and not a permutation, order does not matter.

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R(x) = (500 + ax) (50 - bx)

o-na [289]

Answer:

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500

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250000

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Step-by-step explanation:

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3 years ago

The rate at which a plant grows is constant. At the 3rd week it is 12 inches tall and at the 5th week it is 20 inches tall. If t

Likurg_2 [28]

It would be (12,20) or (20,12) I don't know which one

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3 years ago

Find the volume of the following solid figure. A rectangular solid has sides of 10.5 cm, 6.5 cm, and 8.5 cm. What is its volume?

Zolol [24]

580.125 cm3 is the answer.

6 0

3 years ago

Read 2 more answers

Each statement describes a transformation of the graph of y = lnx. Which statement correctly describes the graph of y = ln(x - 7

vodomira [7]

Hello!

The parent function, y = ln(x), has a vertical and horizontal translation.

y = ln(x - h) + k | In this equation, h is the vertical shift, and k is the horizontal shift.

If ln(x - k), then the graph is translated right k units.

If ln(x + k), then the graph is translated left k units.

If ln(x) + h, then the graph is translated up h units.

If ln(x) - h, then the graph is translated down h units.

Therefore, the graph of y = ln(x - 7) + 3 is translated 3 units up and 7 units to the right, which is choice D.

7 0

3 years ago

Determine the current through each of the LEDs in the circuits below. Which LED will be

Anika [276]

a. $I = 6. 1$ × $10^-4$ A

b. $I = 2. 6$ × $10^-3$ A

c. $I = 0. 04$ A

The LED which would glow brightest is LED C with the greatest current and voltage

The LED which would be the most dim is LED B with low voltage and consequently low current.

<h3>How to determine the current</h3>

The formula for finding current

I = V/R

Where v = voltage

R = resistance

A. V = 12V

R = 4. 7 + 15 = 19. 7 kΩ = 19700 Ω in series

$I = \frac{12}{19700}$

$I = 6. 1$ × $10^-4$ A

B. V = 9V

R = 4. 7 + 1 = 4. 7 kΩ = 4700Ω in series

$I = \frac{12}{4700}$

$I = 2. 6$ × $10^-3$ A

C. V= 12V

1/R = $\frac{1}{750} + \frac{1}{1200} + \frac{1}{950 }$ = $3. 22$ × $10^-3$

R = $\frac{1}{3. 22 * 10 ^-3}$ = 310. 56 Ω

$I = \frac{12}{310. 56}$

$I = 0. 04$ A

It is important to note that the brightness of a bulb depends on both current and voltage depending on whether the bulb it is in parallel or series.

The LED which would glow brightest is LED C with the greatest current and voltage

The LED which would be the most dim is LED B with low voltage and consequently low current.

Learn more about Ohms law here:

brainly.com/question/14296509

#SPJ1

8 0

1 year ago