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Gwar [14]
2 years ago
10

*please help, will thank*

Mathematics
1 answer:
Wittaler [7]2 years ago
7 0
I think the answer is (d) appraisal fee. Hope this helps!
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In addition, from the response shown, using a graphical calculator brings the following benefits:
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 3) You can write the linear equations in any way. Resolving by hand you should probably rewrite the system of equations to find the solution.
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The radioisotope Mo-91 has a t1/2 of 15.5min. A sample decays at the rate of 954counts/min (954cpm). After how many minutes will
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2 years ago
What can you say about the end behavior of the function f(x)=log10(5x-1)
natima [27]
The given function is
f(x) = log₁₀(5x-1)

As x -> -∞, the argument of the log function becomes a large negative number.
Because the log of a negative number is undefined, f(x) is undefined as x -> -∞.

As x -> +∞, the argument of the log function becomes a large positive number.
Therefore f(x) -> +∞ as x -> +∞.

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6 0
2 years ago
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If mJI = (3x+2)°, mHLK = (15x-36)°, and m∠HML = (8x-1)°, find mHLK
Gala2k [10]

Answer:

The value of mHLK will be "(204)°".

Step-by-step explanation:

The given values are:

mJI = (3x+2)°

mHLK = (15x-36)°

and,

m∠HML = (8x-1)°

then,

mHLK = ?

Now,

By using chord-chord formula of angle, we get

mHMK=\frac{1}{2}(mJL+mHLK)

On putting the values in the above formula, we get

⇒  (8x-1)=\frac{1}{2}(15x-36+3x+2)

On applying cross-multiplication, we get

⇒  2(8x-1)=18x-34

⇒  16x-2=18x-34

On subtracting "18x" from both sides, we get

⇒  16x-2-18x=18x-34-18x

⇒  -2x-2=-34

On adding "2" both sides, we get

⇒  -2x=-34+2

⇒  -2x=-32

⇒  x=\frac{32}{2}

⇒  x=16

On putting the value of "x" in mHLK = (15x-36)°, we get

⇒ (15x-36)° = (15×16-36)°

⇒                = (240-36)°

⇒                = (204)°

So that mHLK = (204)°

4 0
2 years ago
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