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yuradex [85]
3 years ago
10

The lunch lady has 6 pounds of lasagna left over. If she makes 1/2-pound servings, how many servings of lasagna can she serve wi

th the amount left over ?
Mathematics
1 answer:
SVEN [57.7K]3 years ago
6 0

Answer: 12 servings

Step-by-step explanation:

If the lunch lady makes equal servings of 1/2 pound each and she had 6 pounds total it would be written like this: 6÷1/2

but you can't divide by a fraction so you multiply by the opposite reciprocal like this 6×2/1 or just 6×2

which gives you your answer of 12

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The answer is D) $0.15

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$2.65 ÷ 5 = 0.15
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Erasers

Based of James, Becky spent a dollar on. 2 boxes of markers
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3 0
3 years ago
Susan estimated she would spend $250 on school clothes at the start of the school year. She actually ended up spending only $230
Anastasy [175]

Answer:

8.70%

Step-by-step explanation:

Percentage error = (difference between estimated value and actual value / actual amount spent) x 100

difference between estimated value and actual value = estimated value - actual amount

$250 - $230 = $20

($20 / $230) x 100 = 8.6957%

4 0
3 years ago
Six ounces is what percent of a pound?
baherus [9]

Answer:

37.5%

Step-by-step explanation:

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8 0
3 years ago
I need help . i’m now in 7th grade and i don’t understand what x and y’s are .
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5 0
3 years ago
Read 2 more answers
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
2 years ago
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