Solve x 2 + 2x

If log a = x and log b = y, then log (ab2) equals

mylen [45]

Answer:

The answer for log(ab²) is x + 2y.

Step-by-step explanation:

You have to apply Logarithm Law,

$log(a) + log(b) \: ⇒ \: log(a \times b)$

$log( {a}^{n} ) \: ⇒ \: n log(a)$

In this question, you have to seperate it out :

$log(a {b}^{2} ) = log(a) + log( {b}^{2} )$

$log( {b}^{2} ) = 2 log(b)$

$let \: log(a) = x$

$let log(b) = y$

$log(a {b}^{2} ) = log(a) + 2 log(b)$

$log(a {b}^{2} ) = x + 2y$

3 0

3 years ago

What is the coordinates of the center of an ellipse defined by the equation 16x^2 + 25y^2 + 160x - 200y + 400 = 0 ? Please give

pickupchik [31]

16x^2 + 25y^2 + 160x - 200y + 400 = 0     Rearrange and regroup.

(16x^2 + 160x) + (25y^2 - 200y ) = 0-400.     Group the xs together and the ys together.

16(X^2 + 10x) + 25(y^2-8y) = -400.     Factorising.

We are going to use completing the square method.

Coefficient of x in the first expression = 10.

Half of it = 1/2 * 10 = 5. (Note this value)

Square it = 5^2 = 25.     (Note this value)

Coefficient of y in the second expression = -8.

Half of it = 1/2 * -8 = -4. (Note this value)

Square it = (-4)^2 = 16. (Note this value)

We are going to carry out a manipulation of completing the square with the values

25 and 16. By adding and substracting it.

16(X^2 + 10x) + 25(y^2-8y) = -400

16(X^2 + 10x + 25 -25) + 25(y^2-8y + 16 -16) = -400

Note that +25 - 25 = 0.    +16 -16 = 0. So the equation is not altered.

16(X^2 + 10x + 25) -16(25) + 25(y^2-8y + 16) -25(16) = -400

16(X^2 + 10x + 25) + 25(y^2-8y + 16) = -400 +16(25) + 25(16)    Transferring the terms -16(25) and -25(16)

to other side of equation. And 16*25 = 400

16(X^2 + 10x + 25) + 25(y^2-8y + 16) = 25(16)

16(X^2 + 10x + 25) + 25(y^2-8y + 16) = 400

We now complete the square by using the value when coefficient was halved.

16(x-5)^2 + 25(y-4)^2 = 400

Divide both sides of the equation by 400

(16(x-5)^2)/400 + (25(y-4)^2)/400 = 400/400              Note also that, 16*25 = 400.

((x-5)^2)/25 + ((y-4)^2)/16 = 1

((x-5)^2)/(5^2) + ((y-4)^2)/(4^2) = 1

Comparing to the general format of an ellipse.

((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) = 1

Coordinates of the center = (h,k).

Comparing   with above   (x-5) = (x - h) , h = 5.

Comparing   with above   (y-k) = (y - k) , k = 4.

Therefore center = (h,k) = (5,4).

Sorry the answer came a little late. Cheers.

3 0

3 years ago

(-11r+14+9r^3) + (3-8r^5+14r^3)

murzikaleks [220]

-8r^5+23r^3-11r+17

Wasn’t Sure What It Was Asking So Hopefully That Works For You

3 0

4 years ago

If the point x,sqrt(3)/2 is on the unit circle. What is x?

sattari [20]

Check the picture below.

if you check your Unit Circle, it'd be there.