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3 years ago

What is the solution to 2log(x)=

1 answer:

4 0

you can't really solve for it unless it equals something or there's a value for x. Assuming this logarithm is base 10... you can set it equal to y and simplify it.

2log(x) = log(x^2)
log(x^2) = y
x^2 = 10^y

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A merchant mixed 12 lb of a cinnamon tea with 3 lb of spice tea. The 15-pound mixture cost $24. A second mixture included 14 lb

bezimeni [28]

Hey there!!

Let us take the price of cinnamon tea as ' x ' and spice tea as ' y '

Now, let us take it into equation

12x + 3y = 15 -------------- ( 1 )

14x + 12y = 45 --------------- ( 2 )

Let us multiply the first equation with 4

48x + 12y = 60 --------------- ( 1 )

14x + 12y = 45 --------------- ( 2 )

Now, subtract equation 2

48x + 12y = 60

-14x - 12y = -45

Now, we get,

34x = 15

x = 15 / 34

This is the cost for cinnamon tea = $ 15 / 34

Now, substitute this into one equation

Let us take this equation - 12x + 3y = 15

Now, substitute ' x '

12 ( 15 / 34 ) + 3y = 15

6 ( 15 / 17 ) + 3y = 15

90 / 17 + 3y = 15

3y = 15 - 90/17

3y = 165 / 17

y = ( 165/17 ) ( 1 / 3 )

y = $ 55/17

Cinnamon tea = $ 15/34

Spice tea = $ 55/17

Hope helps!

3 0

3 years ago

HELP PRECALC NEED IN PROOF FORM

Vinil7 [7]

Hello, please consider the following.

We know the following, right ?

$(\forall a, b \in \mathbb{R}) \left( sin(a+b)=sin(a)sin(b)+cos(a)cos(b) \right)$

So, here, it gives.

$Asin(\omega t+\phi)=Asin(\phi){\sf \bf sin(\omega t)}+Acos(\phi){\sf \bf cos(\omega t)}\\\\=c_2{\sf \bf sin(\omega t)}+c_1{\sf \bf cos(\omega t)}\\\\\text{ *** where }c_2=Asin(\phi) \text{ and } c_1=Acos(\phi) \text{ ***}$

Do not hesitate if you need further explanation.

5 0

4 years ago

Write equation of these in point slope form.

Rufina [12.5K]

1. m=2/5 point= (3,9)

y - 9 = 2/5(x - 3)

2. m= -5 point= (-6,8)

y - 8 = -5(x + 6)

3. m= 0.5 point= (-3,-3)

y + 3 = 0.5(x + 3)

4. m= -4 point= (1,2)

y - 2 = -4(x -1)

3 0

4 years ago

Line ℓ1 has the equation y=x–5 and line ℓ2 has the equation y=x–1. Find the distance between ℓ1 and ℓ2. round your answer to the

Rus_ich [418]

$\ell_1$ and $\ell_2$ are parallel to one another with slope 1. Any line with slope -1 is perpendicular to both of these lines, the simplest being $y=-x$ .