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2 years ago

I need a lot of help, please! ):

Mathematics

1 answer:

tankabanditka [31]2 years ago

8 0

Answer:

I don't know sorry it even very hard and diffult for a high school kid

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Alona [7]

Soln

here

weight = 30 kg

Force = weight × 10

= 30×10

= 300

Area = cm

= 100/100

= 1

now,

Pressure = Force/Area

= 300/1

= 300 pa.

3 0

2 years ago

. A man bought a car for $60 320. After using it for 2 years he decides to trade in the car. The company estimated a depreciatio

Veronika [31]

Answer:

(a) 39,208

(b) 65%

Step-by-step explanation:

(a) The car loses 10% the first year and 15% the second year so that’s 35% lost in total after two years.

35% of 60,320 is 21,112

60,320 - 21,112 = 39,208

39,208 ÷ 60,320 = 0.65 (65%)

8 0

2 years ago

A 4.5 kg cat requires medication at a dose rate of 5 mg/kg once daily for the next 7 days. The tablets are available in 10 mg st

Marta_Voda [28]

Answer:

Approximately 16 tablets of medicine

Step-by-step explanation:

Based on the information above , we can use a Rule of Three to first figure out how many mg the cat would need to ingest per day.

$\frac{5mg}{1kg} = \frac{x}{4.5kg}$

$\frac{5mg*4.5kg}{1kg} = 22.5mg$

Now we see that the cat needs to take a daily dose of 22.5mg. Since the Tablets come at a strength of 10 mg per tablet then,

$22.5 / 10 = 2.25$

The cat would need to take 2.25 tablets per day. Since she needs to take the dose for the next 7 days then,

$2.25 * 7 = 15.75$

She would need to take approximately 16 tablets of medicine during the week .

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

5 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

morpeh [17]

Answer:

kayla and 9968

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A population of 200 animals is decreasing by a rate of .96 per year . At this rate, in how many years will the population be les

mojhsa [17]

At the end of the zeroth year, the population is 200.
At the end of the first year, the population is 200(0.96)¹
At the end of the second year, the population is 200(0.96)²
We can generalise this to become at the end of the nth year as 200(0.96)ⁿ

Now, we need to know when the population will be less than 170.

So, 170 ≤ 200(0.96)ⁿ
170/200 ≤ 0.96ⁿ
17/20 ≤ 0.96ⁿ
Let 17/20 = 0.96ⁿ, first.
log_0.96(17/2) = n

n = ln(17/20)/ln(0.96)
n will be the 4th year, as after the third year, the population reaches ≈176

7 0

3 years ago