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alina1380 [7]
3 years ago
11

An electronics store buys a television at a wholesale price of $120. The store then sells the television to its customers for $3

00. What percent of the wholesale price is the selling price?
Mathematics
1 answer:
algol [13]3 years ago
7 0
180% from the selling price to the wholesale
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Find the area of the figure below
Serhud [2]
Easy

the easiest way is to imagine that you have a rectangle with a corner cut off
the area of the figure=area of rectangle-area cut off

area that was cut off was a triangle

area of rectangle=length times width
area of triangle=1/2base times height

so we draw imaginary lines like in attachment
find area of rectangle
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the top part is ?+5=9, so the base of triangle=4
the side is ?+4=9, so the base is 5
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fgiga [73]

Answer:

y/mz

Step-by-step explanation:

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Show tan(???? − ????) = tan(????)−tan(????) / 1+tan(????) tan(????)<br> .
anyanavicka [17]

Answer:

See the proof below

Step-by-step explanation:

For this case we need to proof the following identity:

tan(x-y) = \frac{tan(x) -tan(y)}{1+ tan(x) tan(y)}

We need to begin with the definition of tangent:

tan (x) =\frac{sin(x)}{cos(x)}

So we can replace into our formula and we got:

tan(x-y) = \frac{sin(x-y)}{cos(x-y)}   (1)

We have the following identities useful for this case:

sin(a-b) = sin(a) cos(b) - sin(b) cos(a)

cos(a-b) = cos(a) cos(b) + sin (a) sin(b)

If we apply the identities into our equation (1) we got:

tan(x-y) = \frac{sin(x) cos(y) - sin(y) cos(x)}{sin(x) sin(y) + cos(x) cos(y)}   (2)

Now we can divide the numerator and denominato from expression (2) by \frac{1}{cos(x) cos(y)} and we got this:

tan(x-y) = \frac{\frac{sin(x) cos(y)}{cos(x) cos(y)} - \frac{sin(y) cos(x)}{cos(x) cos(y)}}{\frac{sin(x) sin(y)}{cos(x) cos(y)} +\frac{cos(x) cos(y)}{cos(x) cos(y)}}

And simplifying we got:

tan(x-y) = \frac{tan(x) -tan(y)}{1+ tan(x) tan(y)}

And this identity is satisfied for all:

(x-y) \neq \frac{\pi}{2} +n\pi

8 0
2 years ago
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barxatty [35]

Answer:

Step-by-step explanation:

Given that,

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dalvyx [7]

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