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3 years ago

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.

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

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.

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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

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3 years ago

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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]

Answer:

Step-by-step explanation:

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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 $\log_e e^a =a$ ]

$\log_e (0.2)=-0.1386t$

$-1.6094379124341 = -0.1386t$

[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

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3 years ago