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raketka [301]
3 years ago
7

Please help I’m confused!

Mathematics
2 answers:
mojhsa [17]3 years ago
8 0

Answer:

2nd option; c=9-a+b

Step-by-step explanation:

  1. The first thing you need to do, is rewrite the equation without the square root.
  2. So for instance \sqrt{16}=4 and 16=(4)^2 therefore a-b+c=(3)^2
  3. Then a-b+c=9
  4. Now we take away ' a ' on both sides and add ' b ' on both sides
  5. So a-(a)-b+(b)+c=9-(a)+(b)
  6. Which gives us 0+0+c=9-a+b which is simply c=9-a+b

Hope this helps

Gnesinka [82]3 years ago
5 0
Number 2 is the answer););!;!:!;!;!;!;!;!;!;!;!;?;
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2 years ago
How can i solve this using a flow chart?
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3 years ago
The length of a rectangular garden is 5 m longer than it's breadth. If it's perimeter is 150 m, find the area of the garden.​
jonny [76]

Answer:

Step-by-step explanation:

L = B + 5

2L + 2B = 150

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3 years ago
Technetium-99m is used as a radioactive tracer for certain medical tests. It has a half-life of 1 day. Consider the function T w
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Answer:

The expression that represents the number of days until only 10% remains is T((d) 10 %) =100×(\frac{1}{2} )^{3.322}.

Step-by-step explanation:

The equation for half life is of the form

A = A₀×(\frac{1}{2} )^{\frac{t}{h} }.........................................................................(1)

Where

A = Final amount

A₀ = Initial amount

t = Time

h = Half life

For the equation T(d) = 100×2⁽⁻²⁾....................................(2)

We have by comparison with the equation for half life

2 ≡ \frac{t}{h}  and and the equation (2) can be written as

Percentage remaining after 2 half lives is

\frac{A}{A_0} ×100=100×(\frac{1}{2} )^{2 }

However if the half life of Technetium-99m is 6 hours then we have for one day

\frac{A}{A_0} ×100=100×(\frac{1}{2} )^{2 *2}

Therefore an expression that represents the number of days until only 10% remains is

\frac{A}{A_0} ×100=100×(\frac{1}{2} )^{\frac{d}{h}  } = 10 %

(\frac{1}{2} )^{\frac{d}{h}  } =\frac{1}{10}

= ㏑(\frac{1}{2} )^{\frac{d}{h}  }  = ㏑(\frac{1}{10})

= \frac{d}{h}×㏑(\frac{1}{2} ) = ㏑(\frac{1}{10})

\frac{d}{h} = \frac{ln(\frac{1}{10}) }{ln(\frac{1}{2} )} = 3.322

Therefore the expression for the number of days 10 % of Technetium-99m will be remaining is

T((d) 10 %) =100×(\frac{1}{2} )^{3.322}

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