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3 years ago

Which would most likely interact with an open flame and cause an accident

Biology

2 answers:

yKpoI14uk [10]3 years ago

3 0

Answer:

D, loose hair.

Explanation:

If your hair is long enough, it could easily drift into the flame and catch on fire.

seropon [69]3 years ago

3 0

Answer:

D. Loose hair

Explanation:

Loose hair would be the most likely to cause an accident with an open flame.

Especially if the hair is long, it can dip down into the flame and catch on fire.

The other objects would not be interacting with the flame because they would be far away, while loose hair can reach into the area where the flame is.

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D. communication

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Neurons are specialised cells that have dendrites, axons and terminal buttons that sends and receives stimuli from the environment and transduces it into a meaningful information and understand the complexities which the brain now functions.

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3 years ago

Eastern Washington, east of the Cascades, has a relatively ____________ climate.

Charra [1.4K]

The answer is dense

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4 years ago

What is TRUE about the relationship between cells and the organism that they are a part of?

Kruka [31]

Answer:

Its A ive done it before if you need proof here http://assessment.aaas.org/items/CE056001#/0

Explanation:

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3 years ago

Read 2 more answers

Why should a female butterfly choose a vibrantly colored male for mating?

denis-greek [22]

Answer:

It attracts the male butterfly

Explanation:

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3 years ago

Bob, a nutritionist who works for the university medical center, has been asked to prepare special diets for two patients, susan

abruzzese [7]

Answer and Explanation:

Possibly, you forgot to attach the table with the contents of each food. Therefore, I calculated the task with similar values.

1. Assume you have Food A, B, and C. Each has certain amount of Calcium, Iron, and Vit C.

Let's designate: Calcium → X oz , Iron → Y oz, vit C → Z oz

2. Assume you have following amount of Calcium in each food:

Food A = 30 mg/oz
Food B = 25 mg/oz
Food C = 20 mg/oz

3. Assume you have following amount of Iron in each food:

Food A = 1 mg/oz
Food B = 1 mg/oz
Food C = 2 mg/oz

4. Assume you have following amount of vit C in each food:

Food A = 2 mg/oz
Food B = 5 mg/oz
Food C = 4 mg/oz

Calculations:

→ You have 3 different food and each has Calcium in it. You need exactly 320mg of Calcium per day. Therefore, each food would comprise certain proportion of daily Calcium intake. We can come up with equation: 30X+25Y+20Z = 390 mg

Similarly, for iron and vitamin C, the next equations could be used:

X+Y+2Z=17 - Iron

2X+5Y+4Z=42 - Vit C

→ Now you have a system of equations:

30X+25Y+20Z = 390 (1)

X+Y+2Z=17 (2)

2X+5Y+4Z=42 (3)

→ Look for similarities! (1) and (3) seem to be similar. We can multiply the (3) by -5 to obtain: -10X-25Y-20Z=-210. We add the (1) and modified (3) equations.

30X+25Y+20Z = 390

-10X-25Y-20Z=-210

__________________

20X+0+0=180

20X=180

→ X=9 (oz of Calcium)

→ Now, we can put x=9 into equations above to ease our job:

270+25Y+20Z=390 → 25Y+20Z=390-270=120

9+Y+Z=17 → Y+2Z=17-9=8

18+5Y+4Z=42 → 5Y+4Z=42-18=24

→ Similarly, we can use Eq-s (2) and (3). We can multiply the (2) by -5 to obtain: -5Y-10Z=-40. We add the (3) and modified (2) equations.

-5Y-10Z=-40

5Y+4Z=24

__________________

-6Z=-24

Z=4

→ Z=4 (oz of vit C)

→ Now, we can take X+Y+Z=17 (2) and just put in known values(X=9 and Z=4) to obtain Y:

9+Y+4=17

→ Y=17-9-4=4 (oz of Iron)

That's all! So, if you have table values, you can just re-write the equations and doing it stepwise, as it is shown here, you can manage to calculate every value you need :)

3 0

3 years ago