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VMariaS [17]
3 years ago
11

A(9,2) B(1,6) if (2,k) is equidistant from A and B, find the value of k.

Mathematics
1 answer:
Pani-rosa [81]3 years ago
5 0
P is the point (2,k)
PA = PB
PA = √(49 + (2-k)²) and PB = √(1 + (6 - k)²)
√(49 + (2-k)²) = √(1 + (6 - k)²) => (49 + (2-k)² = (1 + (6 - k)²
=> 49 + 4 - 4k + k² = 1 + 36 - 12k + k² => 8k = 37 - 53 = -16 => k = -2

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kakasveta [241]

Answer:

k - 2e + 9 = 0

k - e - 4 = 0

Step-by-step explanation:

Given that Elena is currently e years  old and Katie is k years old. Now, nine years ago, Katie was twice as old as Elena was.

So, k - 9 = 2(e - 9)

⇒ k - 2e + 9 = 0 .............. (1)

And, in four years, Elena will be as old as Katie is now.

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Therefore, the equations (1) and (2) matches the conditions. (Answer)

4 0
3 years ago
Under certain conditions, the number of bacteria present in a colony is approximated by f(t) = Age 0.023t, where t is in minutes
AleksandrR [38]

Answer:

  • 5 min: 3,029,058
  • 10 min: 3,398,220
  • 60 min: 10,732,234

Step-by-step explanation:

The given function is evaluated by substituting the given values of t. This requires using the exponential function of your calculator with a base of 'e'. Many calculators have that value built in, or have an e^x function (often associated with the Ln function).

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<h3>5 minutes</h3>

The number of bacteria present after 5 minutes is about ...

  f(5) = 2.7×10^6×e^(0.023×5) ≈ 3,029,058

<h3>10 minutes</h3>

The number of bacteria present after 10 minutes is about ...

  f(10) = 2.7×10^6×e^(0.023×10) ≈ 3,398,220

<h3>60 minutes</h3>

The number of bacteria present after 60 minutes is about ...

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8 0
2 years ago
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katrin2010 [14]
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5 0
3 years ago
Read 2 more answers
A 45-45-90 triangle has legs of length​ 3√2
qwelly [4]

Answer:

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the hypotenuse should be 3√2*√2=6

4 0
3 years ago
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Bezzdna [24]

namely, let's rationalize the denominator in the fraction, for which case we'll be using the <u>conjugate</u> of that denominator, so we'll multiply top and bottom by its <u>conjugate</u>.

so the denominator is 5 + i, simply enough, its conjugate is just 5 - i, recall that same/same = 1, thus (5-i)/(5-i) = 1, and any expression multiplied by 1 is just itself, so we're not really changing the fraction per se.

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4 0
3 years ago
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