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pashok25 [27]
3 years ago
13

What is an equivalent expression to a^3b^2x^4a

Mathematics
1 answer:
lara31 [8.8K]3 years ago
6 0

Answer: a^4 b^2 x^4

Step-by-step explanation:

The exponent "product rule" tells us you, when multiplying two powers that have the same base, you can add the exponents.

so

a^3 b^2 x^4 a^1  

a^3  and a^1 have the same base of a.  So, add 3 + 1 for 4

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Help!!!? Does anyone know the answer to this question!!?
Likurg_2 [28]

Answer:

15.75

36

15.75

Step-by-step explanation:

Formula

the relationship between these two secants is NQ * NP = NM * LM

Givens

  • NP = 32
  • NQ = x
  • NM = 14
  • NL = 14 + 22 = 36

Solution

x*32 = 14*36                 Substitute values. Divide by 32

x = 14 * 36/32               Multiply 14 and 36    

x = 504/32                   Divide by 32

x = 15.75

The answer to the first box is 15.75 or x

The answer to the second box is

LM is assumed to be 22 So LN = MN + LM = 22 + 14= 36

QN = 15.75 is the third answer.


3 0
4 years ago
What is the value of 12x+y(3-23)/5<br> when x = -5 and y = 7?
Anna007 [38]

Answer:=12x−4y

tell me if im wrong

3 0
3 years ago
Read 2 more answers
What's the area of this
Ira Lisetskai [31]
The answer would be 16.5
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4 years ago
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Can someone plz help me?
Oksanka [162]

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7 0
3 years ago
The antibiotic clarithromycin is eliminated from the body according to the formula A(t) = 500e−0.1386t, where A is the amount re
PolarNik [594]

Answer:

Time(t) = 11.61 hours (Rounded to two decimal place)

Step-by-step explanation:

Given: The antibiotic  clarithromycin is eliminated from the body according to the formula:

A(t) = 500e^{-0.1386t}                 ......[1]

where;

A - Amount remaining in the body(in milligram)

t - time in hours after the drug reaches peak concentration.

Given: Amount of drug in the body is reduced to 100 milligrams.

then,

Substitute the value of A = 100 milligrams in [1] we get;

100= 500e^{-0.1386t}

Divide both sides by 500 we get;

\frac{100}{500}=\frac{ 500e^{-0.1386t}}{500}

Simplify:

\frac{1}{5} = e^{-0.1386t}

Taking logarithm both sides with base e, then we have;

\log_e (\frac{1}{5})= \log_e (e^{-0.1386t})

\log_e (\frac{1}{5})=-0.1386t         [ Using \log_e e^a =a ]

or

\log_e (0.2)=-0.1386t

-1.6094379124341 = -0.1386t

 [using value of \log_e (0.2) = -1.6094379124341 ]

then;

t = \frac{-1.6094379124341}{-0.1386}

Simplify:

t ≈11.61 hours.

Therefore, the time 11.61 hours(Rounded two decimal place) will pass before the amount of drug in the body is reduced to 100 milligrams


6 0
3 years ago
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