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Masteriza [31]
3 years ago
15

Cound someone please help me on this, I'm stuck. ​

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
5 0

Negative exponents work like this:

a^{-b}=\dfrac{1}{a^b}

So, in order to evaluate a negative exponent, you simply have to invert the base, and then raise to the positive equivalent of the exponent.

As an example, here are the first three exercises:

8^{-3}=\dfrac{1}{8^3}=\dfrac{1}{512}

(-4)^{-5}=\dfrac{1}{(-4)^5}=-\dfrac{1}{1024}

2k^{-4}=\dfrac{2}{k^4}

You can work out the rest applying this logic.

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There are 12 books on a shelf 5 of these books are new the rest are used what is the ratio of new books to used books
tiny-mole [99]

Answer:

5:7

Step-by-step explanation:

Basically you first find the amount of the books are the shelf knowing that you have 5 new books, you subtract the amount of new books from total books. And 5:7 can't be simplified any further making 5:7 the answer.

5 0
3 years ago
Read 2 more answers
How many pints are in One 11/4 miles
Lostsunrise [7]

3.303e+13 that is approximately

7 0
3 years ago
3. An exclusive clothing boutique triples the price of the items it purchases
Juli2301 [7.4K]

Answer:

a) MARKUP PERCENT = 200%

b) <u>If the cost price of clothing A  = $10</u>

    then its retail price  = $30

   <u> If the cost price of clothing B   = $25</u>

   then its retail price  = $75

Step-by-step explanation:

Let us assume the original price for the clothing = a

Now,  boutique triples the price of the items it purchases for resale.

So, the resale value of the item of clothing = 3 times (Original price) = 3 a

So, here: The resale Value  = 3 a

a)  MARK UP Percentage = (\frac{\textrm{SP - CP}}{CP})  \times 100

\implies \%  = (\frac{3a-a}{a} ) \times 100 =  200 \%

or, the MARKUP PERCENT = 200%

b) Two expressions that represent retail price

<u>If the cost price of clothing A  = $10</u>

then its retail price  = 200% of 10 +  10  = 20 + 10 = $30

<u>If the cost price of clothing B   = $25</u>

then its retail price  = 200% of 25 +  25  = 50 + 25 = $75

7 0
3 years ago
Find the equation of a straight line passing through the point (0,1) which is perpendicular to the line y=-2x+2
alexandr1967 [171]

Answer:

y = \frac{1}{2} x + 1

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - 2x + 2 ← is in slope- intercept form

with slope m = - 2

Given a line with slope m then the slope of a line perpendicular to it is

m_{perpendicular} = - \frac{1}{m} = - \frac{1}{-2} = \frac{1}{2}

The line crosses the y- axis at (0, 1) ⇒ c = 1 , then

y = \frac{1}{2} x + 1 ← equation of perpendicular line

3 0
2 years ago
Uninhibited growth can be modeled by exponential functions other than​ A(t) ​=Upper A 0 e Superscript kt. For ​ example, if an i
laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than A(t)=A_{0}e^{kt}. for example, if an initial population P₀ requires n units of time to triple, then the function P(t)=P_{0}(3)^{\frac{t}{n} } models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form P(t)=P_{0}(3)^{\frac{t}{n} } that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form A(t)=A_{0}e^{kt}.

Answer: a) P(t)=50(3)^{\frac{t}{30} }

              b) P(t) = 280 insects

              c) t = 74 days

             d) A(t)=50e^{0.037t}

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

P(t)=50(3)^{\frac{t}{30} }

b) In t = 47 days:

P(t)=50(3)^{\frac{t}{30} }

P(47)=50(3)^{\frac{47}{30} }

P(47)=50(3)^{1.567}

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

750=50(3)^{\frac{t}{30} }

\frac{750}{50}=(3)^{\frac{t}{30} }

(3)^{\frac{t}{n} }=15

Using the property <u>Power</u> <u>Rule</u> of logarithm:

log(3)^{\frac{t}{30} }=log15

\frac{t}{30}log(3)=log15

t=\frac{log15}{log3} .30

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

A(t)=A_{0}e^{kt}

3A_{0}=A_{0}e^{30k}

3=e^{30k}

Use Power Rule again:

ln3=ln(e^{30k})

ln3=30k

k=\frac{ln3}{30}

k = 0.037

Equation for exponential growth will be:

A(t)=50e^{0.037t}

3 0
3 years ago
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