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2 years ago

7

you are planning to purchase displays for your store which has 2400 sq ft of selling space. Each display has a footprint of 10 f

t x 4 ft. You want to leave 60% open space in your store for walking around. How many displays can you fit?

1 answer:

Fynjy0 [20]2 years ago

5 0

60% of 2,400 is 1,440. So in all actuallity, you only have 960 square feet to fill with displays. 10*4= 40ft^2. 960/40=24.

You can fit 24 displays in your store while having 60% walking space.

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Answer:

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Step-by-step explanation:

We have the fraction $\frac{2m}{2m+3} -\frac{2m}{2m-3}$

Step 1. Use LCM of the fraction, (2m+3)(2m-3), to simplify the fraction:

$\frac{(2m-3)(2m)-(2m+3)(2m)}{(2m+3)(2m-3)} =\frac{2m^2-6m-[2m^2+6m]}{(2m+3)(2m-3)} =\frac{2m^2-6m-2m^2-6m}{(2m+3)(2m-3)} =\frac{-12m}{(2m+3)(2m-3)}$

Step 2. Equate the resulting fraction to zero and solve for $m$ :

$\frac{-12m}{(2m+3)(2m-3)} =0$

$-12m=0[(2m+3)(2m-3)]$

$-12m=0$

$m=\frac{0}{-12}$

$m=0$

Step 3. Replace the value in the original equation and check if it holds:

$\frac{-12m}{(2m+3)(2m-3)} =0$

$-12m=0$

Since $m=0$ ,

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Since the only solution of the equation holds, the equation bellow doesn't have any extraneous solution

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Answer:

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Javier is making a cake for a party. The recipe calls for 2 3 4 234 cups of flour, but he wants to triple the recipe. He guessed

GalinKa [24]

The correct question is
<span>Javier is making a cake for a party. The recipe calls for 2 3/4 cups of flour, but he wants to triple the recipe. He guessed and put in 7 cups of flour. How many more cups of flour should he have used?

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the answer is
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2 years ago

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lidiya [134]

Answer:

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And replacing we got:

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For this case w eknow the following parameters:

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We want to find the following probability:

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We can use the z score formula given by:

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And replacing we got:

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And using the normal standard distribution and the complement rule we got:

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natali 33 [55]

Answer:

a) (see attached picture)

b) d, h, velocity of helicopter, dd/dt

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Step-by-step explanation:

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$d^{2}=h^{2}+x^{2}$

we can use this equation to find the value of x at that very time, so we can do that by solving the equation for x, so we get:

$x=\sqrt{d^{2}-h^{2}}$

and substitute the given values:

$x=\sqrt{(1)^{2}-(0.5)^{2}}$

which yields:

x=0.866mi

we know that the height of the helicopter is going to be constant, so we rewrite the equation as:

$d^{2}=(0.5)^{2}+x^{2}$

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Next, we can go ahead and differentiate the equation so we get:

$2d\frac{dd}{dt}=2x\frac{dx}{dt}$

which can be simplified to:

$d\frac{dd}{dt}=x\frac{dx}{dt}$

we can next solve the equation for dx/dt so we get:

$\frac{dx}{dt}=\frac{d}{x}*\frac{dd}{dt}$

so we can now substitue te provided values so we get:

$\frac{dx}{dt}=\frac{1}{0.866}*-190$

so we get:

$\frac{dx}{dt}=-219.40mph$

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now, this problem deals with relative velocities, so we get that:

Velocity of car about the helicopter = Velocity of the car - Velocity of the helicopter. Or:

$V_{ch}=V_{c}-V_{h}$

so we can solve this for the actual velocity of the car, so we get:

$V_{c}=V_{ch}+V_{h}$

so we get:

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