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wolverine [178]
1 year ago
9

in a sale the cost of a coat reduced from $85 to $67.50 calculate the percentage reduction in the cost of the coat

Mathematics
1 answer:
ludmilkaskok [199]1 year ago
8 0
20.58% would be the answer
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Need help<br> Question in the screenshot below<br> NO spam<br> Please show your work
Alja [10]
<h3>Answer:   x = 6</h3>

======================================================

Work Shown:

\log_{4}(x+10)+\log_{4}(x-2)=\log_{4}(64)\\\\\log_{4}\left((x+10)(x-2)\right)=\log_{4}(64)\\\\(x+10)(x-2)=64\\\\x^2-2x+10x-20=64\\\\x^2-2x+10x-20-64=0\\\\

x^2+8x-84=0\\\\(x+14)(x-6)=0\\\\x+14=0 \ \text{ or } \ x-6=0\\\\x=-14 \ \text{ or } \ x=6\\\\

Those are the possible solutions, but plugging x = -14 back into the original equation will lead to an error. So we rule x = -14 out

x = 6 works as a solution however

3 0
3 years ago
The momentum of an object, p, is determined by its mass, m, and its velocity, v. The formula for this relationship is shown.
timama [110]

Answer:

b

Step-by-step explanation:

7 0
3 years ago
Need help please for school thank uuuuu
kramer

Answer:

>

Step-by-step explanation:

4 0
2 years ago
Hello im in 11th grade, im currently struggling in two classes: chemistry and spanish. Im already in the second semester, and ju
Katarina [22]
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7 0
3 years ago
A piece of cardboard is 15 inches by 30 inches. A square is to be cut from each corner and the sides folded up to make an open-t
solmaris [256]

Answer:

Maximum volume = 649.519 cubic inches

Step-by-step explanation:

A rectangular piece of cardboard of side 15 inches by 30 inches is cut in such that a square is cut from each corner. Let x be the side of this square cut. When it was folded to make the box the height of box becomes x, length becomes (30-2x) and the width becomes (15-2x).

Volume is given by  

V = V = Length\times Width\times Height\\V = (30 - 2x)(15-2x)x= 4x^3-90x^2+450x\\So,\\V(x) = 4x^3-90x^2+450x

First, we differentiate V(x) with respect to x, to get,

\frac{d(V(x))}{dx} = \frac{d(4x^3-12x^2+9x)}{dx} = 12x^2 - 180x +450

Equating the first derivative to zero, we get,

\frac{d(V(x))}{dx} = 0\\\\12x^2 - 180x +450 = 0

Solving, with the help of quadratic formula, we get,

x = \displaystyle\frac{5(3+\sqrt{3})}{2}, \frac{5(3-\sqrt{3})}{2},

Again differentiation V(x), with respect to x, we get,

\frac{d^2(V(x))}{dx^2} = 24x - 180

At x =

\displaystyle\frac{5(3-\sqrt{3})}{2},

\frac{d^2(V(x))}{dx^2} < 0

Thus, by double derivative test, the maxima occurs at

x = \displaystyle\frac{5(3-\sqrt{3})}{2} for V(x).

Thus, largest volume the box can have occurs when x = \displaystyle\frac{5(3-\sqrt{3})}{2}}.

Maximum volume =

V(\displaystyle\frac{5(3-\sqrt{3})}{2}) = (30 - 2x)(15-2x)x = 649.5191\text{ cubic inches}

8 0
3 years ago
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