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3 years ago

Find an expression equivalent to the one shown below.

Mathematics

1 answer:

TEA [102]3 years ago

6 0

Answer:

C) 1/6^32

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)^c = a^(b·c)

(a^b)(a^c) = a^(b+c)

a^-b = 1/a^b

_____

Using these rules, your expression simplifies to ...

6^8·6^(10(-4)) = 6^8·6^-40 = 6^(8-40) = 6^-32 = 1/6^32

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Solve for x.<br> x-2/4 - x+3/3 = -2

exis [7]

The LCD is 12. Thus, your

(x-2)/4 - (x+3)/3 = -2 becomes 12((x-2)/4 - (x+3)/3 = -2), or

3(x-2) - 4(x+3) = -36

Then 3x - 6 - 4x - 12 = -36. Combining like terms,

-x = -36 + 18, or -x = -18, or x = 18.

Check this result by subbing 18 for x in the given equation.

Please note: You must use parentheses here as indicated to eliminate any ambiguity about order of operations and what divides into what.

7 0

4 years ago

How do you solve this 2x +5 - 7x = 15

Serga [27]

1. combine like terms (2x and -7x) to create -5x
2. subtract 5 from each side to create: -5x=10
3. divide both sides by -5
4. x=-2

7 0

3 years ago

Sophia buys an apple that weighs 0.45 pound, a grapefruit that weighs pound, a navel orange that weighs pound, and a pear that w

Tanzania [10]

Answer:

The order of fruit from least to greatest weight is -

Apple , pear, navel orange, grapefruit

Step-by-step explanation:

P.S - The exact question is -

Given - Sophia buys an apple that weighs 0.45 pound, a grapefruit that weighs $\frac{3}{4}$ pound, a navel orange that weighs $\frac{5}{8}$ pound, and a pear that weighs 0.5 pound.

To find - What is the order of the fruit from least weight to the greatest weight ?

Proof -

Weight of apple = 0.45 pound

Weight of grapefruit = $\frac{3}{4}$ pound = 0.75 pound

Weight of navel orange = $\frac{5}{8}$ pound = 0.625 pound

Weight of a pear = 0.5 pound

As,

0.45 < 0.5 < 0.625 < 0.75

So, The order of fruit from least to greatest weight is -

Apple , pear, navel orange, grapefruit

8 0

3 years ago

The growth of a city is described by the population functionwhereis the initial population of the city, t is the time in years,

mamaluj [8]

Complete question:

The growth of a city is described by the population function p(t) = P0e^kt where P0 is the initial population of the city, t is the time in years, and k is a constant. If the population of the city atis 19,000 and the population of the city atis 23,000, which is the nearest approximation to the population of the city at

Answer:

27,800

Step-by-step explanation:

We need to obtain the initial population(P0) and constant value (k)

Population function : p(t) = P0e^kt

At t = 0, population = 19,000

19,000 = P0e^(k*0)

19,000 = P0 * e^0

19000 = P0 * 1

19000 = P0

Hence, initial population = 19,000

At t = 3; population = 23,000

23,000 = 19000e^(k*3)

23000 = 19000 * e^3k

e^3k = 23000/ 19000

e^3k = 1.2105263

Take the ln

3k = ln(1.2105263)

k = 0.1910552 / 3

k = 0.0636850

At t = 6

p(t) = P0e^kt

p(6) = 19000 * e^(0.0636850 * 6)

P(6) = 19000 * e^0.3821104

P(6) = 19000 * 1.4653739

P(6) = 27842.104

27,800 ( nearest whole number)

3 0

3 years ago

What is StartFraction 5 Over 8 EndFraction divided by One-fourth?

RideAnS [48]

Answer:

1/3

Step-by-step explanation:

8 0

3 years ago

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