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IceJOKER [234]
3 years ago
10

I give a medal if you can figure this out fast! A certain forest covers an area of 4200 km2 . Suppose that each year this area d

ecreases by 8.5% . What will the area be after 8 years?
Mathematics
2 answers:
Mars2501 [29]3 years ago
8 0
The answer would be 525 kilometers2


oee [108]3 years ago
7 0
<span>1541.213 square km would be correct

</span>
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a rectangle has a length 3 meters more than three times the width. the area of the rectangle is less than 90 meters squared. wri
Natasha2012 [34]

Answer: -6<x<5

Step-by-step explanation:

Given

Length is 3 m more than three times the width

Suppose width is x

So, the length is 3x+3

The area of a rectangle is  length\times width

Area= (3x+3)x

\Rightarrow 3x^2+3x-90

6 0
2 years ago
Need help asap pleas and thank you .
Arisa [49]

Answer:

The answer that is <em>not </em>true when x = -4 is <u>B</u>.

Step-by-step explanation:

Since x = -4, that means all we have to do to eliminate answers is plug -4 in place of x.

A is incorrect because 2 × -4 = -8, which means that answer IS true and we're looking for the one that isn't true.

Same with C, -4 + -4 + -4 added together gives you -12 = -12, therefore isn't the right choice.

Finally D. -4/4 divided equals -1, so that would make it -1 = -1.

B is the only one that doesn't make sense, therefore you have your answer.

4 0
2 years ago
2x [(12 x 2) - 5 + 34] help
seraphim [82]

2x((12x2)-5+34)

2x(24-5+34)

(48x-10x+68x)

Distributive property?

116x?

8 0
3 years ago
A desk is on sale for 34%  off. The sale price is $363  .What is the regular price? I tried this problem so many times but I kee
elena-14-01-66 [18.8K]
The answer is the regular price is $486.42
8 0
3 years ago
Read 2 more answers
When an amount of heat Q (in kcal) is added to a unit mass (kg) of a
lisov135 [29]

Answer:

The change in temperature per minute for the sample, dT/dt is 71.\overline {6} °C/min

Step-by-step explanation:

The given parameters of the question are;

The specific heat capacity for glass, dQ/dT = 0.18 (kcal/°C)

The heat transfer rate for 1 kg of glass at 20.0 °C, dQ/dt = 12.9 kcal/min

Given that both dQ/dT and dQ/dt are known, we have;

\dfrac{dQ}{dT} = 0.18 \, (kcal/ ^{\circ} C)

\dfrac{dQ}{dt} = 12.9 \, (kcal/ min)

Therefore, we get;

\dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = {\dfrac{dQ}{dt} } \times \dfrac{dT}{dQ} = \dfrac{dT}{dt}

\dfrac{dT}{dt} = \dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = \dfrac{12.9 \, kcal / min }{0.18 \, kcal/ ^{\circ} C }   = 71.\overline 6 \, ^{\circ } C/min

For the sample, we have the change in temperature per minute, dT/dt, presented as follows;

\dfrac{dT}{dt}  = 71.\overline 6 \, ^{\circ } C/min

8 0
2 years ago
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