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Molodets [167]
4 years ago
5

Please help on this question.

Mathematics
1 answer:
Sonbull [250]4 years ago
6 0

Answer:

C

Step-by-step explanation:

\frac{ - 20 {x}^{2}  + 14x + 12}{5x - 6}  \\  \\  =   \frac{- (20 {x}^{2}   - 14x - 12)}{5x - 6}  \\  \\ =   \frac{ - (5x - 6)(4x + 2)}{5x - 6}  \\  \\  =  - (4x + 2 ) \\  =  - 4x - 2

Domain:

g(x) ≠ 0

5x - 6 = 0 \\ 5x = 6 \\  \\ x =  \frac{6}{5}

All real numbers ≠ 6/5

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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
Alinara [238K]

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}

3 0
4 years ago
Simplify<br>2⁴×2³/2⁵×2⁴<br>leaving your answer in index form​
igomit [66]

Answer:

hope this help

Step-by-step explanation:

2⁴×2³/2⁵×2⁴

= 2^{7}/2^{9}  = 1/2^{2}

7 0
3 years ago
Read 2 more answers
Will give Brainly What is the relationship between the sine and cosine of complementary angles in a right triangle? Please give
Lina20 [59]

Answer:

they are equal

Step-by-step explanation:

Consider the right triangle ABC with sides a, b, and c as shown in the figure.  

Let m(\angle A)=\alpha, and m(\angle B)=\beta.

\alpha +\beta=90^{\circ}, so angles A and B are complementary.

According to the definition of sine, and cosine:

\displaystyle{ \sin \alpha= \frac{opposite\ side}{hypotenuse} =\frac{a}{c} , and

\displaystyle{ \cos \beta= \frac{adjacent\ side}{hypotenuse} =\frac{a}{c}  

4 0
3 years ago
Write an equation that represents “Eight more than the quotient of a number and two is fourteen.”
seropon [69]

Answer:

I think it's supposed to be 8x+2=14

5 0
4 years ago
PLEASE HELP ME OUT BECAUSE I AM ON A TIME CRUNCH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
anyanavicka [17]
A=LW and L=W+6, W=L-6 using this W in the area formula

A=L(L-6)

A=L^2-6L

L^2-6L-A=0  and we are told that area=72 so

L^2-6L-72=0  factor

L^2-12L+6L-72=0

L(L-12)+6(L-12)=0

(L+6)(L-12)=0, since L>0

L=12 meters
6 0
3 years ago
Read 2 more answers
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