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docker41 [41]
3 years ago
9

A store sells ladders. • The retail price was a 40 percent markup over the manufacturer price. • A month later, the store reduce

d the retail price of the ladder by 25 percent. What percent markup is the new retail price over the manufacturer price?
Mathematics
1 answer:
TEA [102]3 years ago
6 0
For this item, we represent the very original value or price of the given item by x such that increasing the retail price by 40 percent will give us 1.4x. Also, when the retail price of the item is reduced by 25 percent, we have.
                          new retail price = 1.4x (1 - 0.25) = 1.05x. 
This means that the new retail price is only 5% over the manufacturer price. 
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The difference of 3 times a number and 15 is no less than the square of the number.
jok3333 [9.3K]

Answer:

(B) The inequality that represents this relationship is 3x  < x^2 +15

Step-by-step explanation:

Let us assume the given number = x

⇒ 3 times the given number = 3 (x)  = 3x

Square of the given number = (x)^2  = x^2

Now, According to the question:

The difference of (3 x) and 15 is no less than x^2

⇒ 3x - 15 < x^2\\ \implies 3x  < x^2 +15

or, 3x  < x^2 +15

Hence, the given inequality is represented as 3x  < x^2 +15

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3 years ago
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Is the answer A or C?
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Answer:

it's C

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1. What are the roots of 4x2 = 8x - 7?
Andrei [34K]
4 times 2 =8times-7 is 1
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Mike observed that 75% of the students of a school liked skating. If 35 students of the school did not like skating, the number
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100-75=25% did not like
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EXAMPLE 5 If F(x, y, z) = 4y2i + (8xy + 4e4z)j + 16ye4zk, find a function f such that ∇f = F. SOLUTION If there is such a functi
Valentin [98]

If there is such a scalar function <em>f</em>, then

\dfrac{\partial f}{\partial x}=4y^2

\dfrac{\partial f}{\partial y}=8xy+4e^{4z}

\dfrac{\partial f}{\partial z}=16ye^{4z}

Integrate both sides of the first equation with respect to <em>x</em> :

f(x,y,z)=4xy^2+g(y,z)

Differentiate both sides with respect to <em>y</em> :

\dfrac{\partial f}{\partial y}=8xy+4e^{4z}=8xy+\dfrac{\partial g}{\partial y}

\implies\dfrac{\partial g}{\partial y}=4e^{4z}

Integrate both sides with respect to <em>y</em> :

g(y,z)=4ye^{4z}+h(z)

Plug this into the equation above with <em>f</em> , then differentiate both sides with respect to <em>z</em> :

f(x,y,z)=4xy^2+4ye^{4z}+h(z)

\dfrac{\partial f}{\partial z}=16ye^{4z}=16ye^{4z}+\dfrac{\mathrm dh}{\mathrm dz}

\implies\dfrac{\mathrm dh}{\mathrm dz}=0

Integrate both sides with respect to <em>z</em> :

h(z)=C

So we end up with

\boxed{f(x,y,z)=4xy^2+4ye^{4z}+C}

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