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3 years ago

What is 47 and 2/3 as a decimal

Mathematics

1 answer:

Phoenix [80]3 years ago

4 0

47 is 0.47 and 2/3 is 0.66

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Question 1

Oliga [24]

Where are the choices

6 0

3 years ago

25y + 150x = 250<br><br> find the Variables

fomenos

Answer: Step-by-step explanation:

25y + 150x = 250 (÷25)

y+6x=10

y=10-6x

example x=1 y=4

8 0

2 years ago

Solve for y<br> -3y+4=-17

patriot [66]

Answer:

Step-by-step explanation:

-3y+4= -17

-3y= -21

y= 7

4 0

3 years ago

Read 2 more answers

4(3x^2y^4)^3 / (2x^3y^5)^4

tatyana61 [14]

Solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Step-by-step explanation:

We need to Solve the expression: $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

Solving:

$\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

$=\frac{4(3^3x^6y^{12})}{(2^4x^{12}y^{20})}\\=\frac{4(27x^6y^{12})}{(16x^{12}y^{20})}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}$

Applying exponent rule: $\frac{x^a}{x^b}=x^{a-b}$

$=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{27x^{-6}y^{-8}}{4}$

Another exponent rule says: $x^{-a}=\frac{1}{x^a}=$

$=\frac{27}{4x^{6}y^{8}}$

So, solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Keywords: Solving Exponents

Learn more about Solving Exponents at:

#learnwithBrainly

8 0

3 years ago

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

3 years ago