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Viefleur [7K]
3 years ago
14

How to evaluate the expression 10(6-5)-3(9-6)

Mathematics
2 answers:
EastWind [94]3 years ago
6 0
<span>10(6-5)-3(9-6) = 10 x 1 - 3 x 3 = 10 - 9 = 1</span>
Levart [38]3 years ago
3 0
You can do it two ways with PEMDAS or with Distributive property.
D property- 10x6=60 10x5=50 60-50=10 then 3x9=27 3x6=18 27-18=9
10-9=1
Pemdas- 6-5=1x10=10 9-6=3x3=9 10-9=1 either way your answer is
1 (one)
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Find the quotient.<br> 52.06 divided by 1.9
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27.4

Step-by-step explanation:

First multiply the numerator and denominator by 10

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Then add a decimal point to the solution

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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

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