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3 years ago

Nike shoes were originally $40 and is now marked up by 5% what is the price after the markup

Mathematics

1 answer:

erastova [34]3 years ago

7 0

Answer : 42

If the shoes were marked up by 5% then:

5 ÷ 100 × 40 = $2

Add the $2 to the $40 and you'll get:

$40 + 2 = $42

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Find all solutions to the equation in the interval [0, 2pi). COS X = sin 2x

Lera25 [3.4K]

Answer:

Step-by-step explanation:

cos x=sin 2x

sin 2x-cos x=0

2 sin x cos x-cos x=0

cos x(2 sin x-1)=0

either cos x=0

$x=\frac{\pi }{2},\frac{3\pi }{2}$

2 sin x-1=0

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2 years ago

Simply the expression by combining like terms 20c+8d+10c-5d

valentinak56 [21]

First, combine 20c and 10c because they are like terms then combine 8d and -5d.

It would look like this:

20c+8d+10c-5d

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Hope that helps

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2 years ago

Write a Let" statement for the unknown, write an equation, and solve the equation. 11. Jane bought a number of records for $7 ea

Advocard [28]

Let n = number of records.

Each record costs $7, so n records cost 7n.

She then spent $3, so the total spent is 3n + 3.

She spent a total of $24, so 3n + 3 must equal 24. That gives us the following equation.

7n + 3 = 24

Subtract 3 from both sides.

7n = 21

Divide both sides by 7.

n = 3

Answer: She bought 3 records.

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3 years ago

When a bactericide is added to a nutrient broth in which bacteria are growing, the bacteria population continues to grow for a

baherus [9]

Answer:

a) 1296 bacteria per hour

b) 0 bacteria per hour

c) -1296 bacteria per hour

Step-by-step explanation:

We are given the following information in the question:

The size of the population at time t is given by:

$b(t) = 9^6 + 6^4t-6^3t^2$

We differentiate the given function.

Thus, the growth rate is given by:

$\displaystyle\frac{db(t)}{dt} = \frac{d}{dt}(9^6 + 6^4t-6^3t^2)\\\\= 6^4-2(6^3)t$

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$\displaystyle\frac{db(t)}{dt} \bigg|_{t=0}= 6^4-2(6^3)(0) = 1296\text{ bacteria per hour}$

b) Growth rates at t = 3 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=3}= 6^4-2(6^3)(3) = 0\text{ bacteria per hour}$

c) Growth rates at t = 6 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=6}= 6^4-2(6^3)(6) = -1296\text{ bacteria per hour}$

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3 years ago

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alexira [117]

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3 years ago