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kirill [66]
3 years ago
7

4. A six-meter-long ladder leans against a building. If the ladder makes an angle of 60° with the

Mathematics
2 answers:
ankoles [38]3 years ago
7 0

Answer: the answer is simply 5.2cm

Step-by-step explanation:

serious [3.7K]3 years ago
5 0
Rather than memorize sines and cosines of various common triangles, I prefer to remember the simple derivation of those formulas…

Imagine that the wall is a mirror, so you have one ladder leaning against the wall and you also have the reflection of that ladder leaning behind it. Now the ladder is 6m long, its reflection is 6m long, and if the distance between their bases is also 6m then you have a 60–60–60 equilateral triangle. And, good news, the problem says that the base angle is 60 degrees. So the real part in front of the mirror is half of the equilateral triangle, which means the distance from the base to the mirror is 3m. And then the height is sqrt(36–9) = sqrt(27) = 3*sqrt(3).

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PLEASE HELP<br><br><br> Solve.<br><br> 5x+2/2 = 2x - 1<br><br> X = ________ ?
lakkis [162]
5x + 2/2 = 2x - 1
2/2 is basically 1 so
5x + 1 = 2x - 1
isolate the variable
5x + 1 = 2x - 1
      -1          -1
5x = 2x -2
-2x   -2x
3x = -2
solve
3x/3=-2/3
x=-2/3
4 0
3 years ago
A population of 30 deer are introduced into a wildlife sanctuary. It is estimated that the sanctuary can sustain 600 deer. The p
Ksju [112]

Complete question :

A population of 30 deer are introduced into a wildlife sanctuary. It is estimated that the sanctuary can sustain 600 deer. The population would grow by 30 percent per year

how many after one year

how many after two years

Answer:

39 deers

51 deers

Step-by-step explanation:

The question can be expressed using the compounding rate formula:

A = P(1+r)^t

A = final population ; P = initial population ; rate, r = 30% = 0.3 ; t = time

After 1 year, t = 1

A = 30(1 + 0.3)^1

A = 30(1.3)^1

A = 39 deers

B.)

After 1 year, t = 2

A = 30(1 + 0.3)^2

A = 30(1.3)^2

A = 50.7

A = 51 deers approximately

4 0
3 years ago
Bismuth-210 has a half-life of about 5 days. After 15 days, how many milligrams of a 600 mg sample will remain?
raketka [301]
Every 5 days the number is cut in half. So from 600 mg to day 5, the sample will be 300 mg. Then from day 5 to day 10, the sample will reduce to 150 mg. And finally, from day 10 to day 15, the sample will have reduced to:

75 mg

I hope this Helps!
5 0
3 years ago
Read 2 more answers
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BabaBlast [244]
The answers are 2 and 5 
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3 years ago
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