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Bogdan [553]
3 years ago
12

A store buys an item for $50 and marks it up 100%. What is the price?

Mathematics
2 answers:
Alexxandr [17]3 years ago
7 0
50 + (100% of 50) =
50 + 1(50) =
50 + 50 =
100 <===
pantera1 [17]3 years ago
3 0

Answer:

Therefore, store has marked the selling price as $100.

Step-by-step explanation:

A store buys an item for $50 and marks it up 100% up means store has marked the selling price of that item twice of its cost price.

Selling price = Cost price + Profit

                     = Cost price + 100% of cost price

                    = 2 × cost price

                     = 2 ×50 = $100

Therefore, store has marked the selling price as $100.

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expeople1 [14]

Answer:

x = 4y + 28

Step-by-step explanation:

x/4 - 7 = y

+7              +7

x/4 = y + 7

× 4             × 4

x = 4y + 28

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Twelve more than half of a number is equal to the number
tester [92]
Let's consider that number to be 'x'..
So, twice the number (2x) is 12 greater than than the half of the number.
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2x=12+x/2
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2x=(24+x)/2
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2 years ago
Triangle PQR is formed by the three squares A, B, and C:
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The first choice if ()2 means to the power of two if not then it is none of them.
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kondor19780726 [428]

The right answer is Option A:x=\frac{rs-t}{2}

Step-by-step explanation:

Given expression is;

s=\frac{2x+t}{r}

We will solve this expression for x, therefore,

Multiplying both sides by r

r*s=\frac{2x+t}{r}*r\\rs=2x+t

Subtracting t from both sides

rs-t=2x+t-t\\rs-t=2x\\2x=rs-t

Dividing both sides by 2

\frac{2x}{2}=\frac{rs-t}{2}\\x=\frac{rs-t}{2}

The right answer is Option A:x=\frac{rs-t}{2}

Keywords: subtraction, division

Learn more about subtraction at:

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7 0
3 years ago
A bacteria culture starts with 400 bacteria and grows at a rate proportional to its size. After 4 hours, there are 9000 bacteria
Kaylis [27]

Answer:

A) The expression for the number of bacteria is P(t) = 400e^{0.7783t}.

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

P' = kP

where P stands for the population of bacteria.

Writing P' as \frac{dP}{dt}, we get

\frac{dP}{dt} = kP.

Notice that this is a separable equation, so

\frac{dP}{P} = kdt.

Then, integrating in both sides of the equality:

\int\frac{dP}{P} = \int kdt.

We have,

\ln P = kt+C.

Now, taking exponential

P(t) = Ce^{kt}.

The next step is to find the value for the constant C. We do this using the initial condition P(0)=400. Recall that this is the initial population of bacteria. So,

400 = P(0) = Ce^{k0}=C.

Hence, the expression becomes

P(t) = 400e^{kt}.

Now, we find the value for k. We are going to use that P(4)=9000. Notice that

9000 = 400e^{k4}.

Then,

\frac{90}{4} = e^{4k}.

Taking logarithm

\ln\frac{90}{4} = 4k, so \frac{1}{4}\ln\frac{90}{4} = k.

So, k=0.7783788273, and approximating to the fourth decimal place we can take k=0.7783. Hence,

P(t) = 400e^{0.7783t}.

B) To find the number of bacteria after 5 hours, we only need to evaluate the expression we have obtained in the previous exercise:

P(5) =400e^{0.7783*5} = 19593.723 \approx 19593.  

C) In this case we want to do the reverse operation: we want to find the value of t such that

30000 = 400e^{0.7783t}.

This expression is equivalent to

75 = e^{0.7783t}.

Now, taking logarithm we have

\ln 75 = 0.7783t.

Finally,

t = \frac{\ln 75}{0.7783} \approx 5.55.

So, after 5.55 hours the population of bacteria will reach 30000.

6 0
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