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netineya [11]
3 years ago
9

Which expression is equivalent?

Mathematics
1 answer:
gregori [183]3 years ago
8 0
The correct option is: Option (C) log_3c - log_{3}9

Explanation:
It is the property of log that if it contains the fraction, it can be expressed as the difference of the log of numerator and the log of denominator.

The general form is:
log( \frac{x}{y} ) = log(x) - log(y)

Hence the correct answer is log_3c - log_{3}9.


You might be interested in
how so you solve 5x+27=9(x+3)-4x and is this a conditional equation, an identity, or an inconsistent equation
Vikentia [17]

To solve this, follow these steps:

5x + 27 = 9 (x+3) - 4x

Distribute the 9 within the parenthesis:

5x + 27 = 9x + 27 - 4x

Then combine like terms:

5x + 27 = 5x + 27

These two then cancel each other out completely.

This is an identity because an identity in math is when both equations equal each other, which is shown above.

Hope this helps!

5 0
3 years ago
Ariana wants to ride her bicycle 37 miles this week. She has already ridden 14 miles. If she rides for 5 more days, write and so
lakkis [162]

Answer:

4.6 miles for 5 days.

Step-by-step explanation:

37 - 14 = 23

23 divided by 5 = 4.6

7 0
3 years ago
19 6/7 divided by 4 1/8
anygoal [31]
<span>4</span>\frac{188}{231}
Hope this helps!
7 0
2 years ago
Denise bought 116 ounces of beans for a bean dip. She bought both 15-ounce cans and 28-ounce cans, and the total number of cans
Valentin [98]
15x + 28y = 116
x + y = 6

that would be ur system of equations

7 0
3 years ago
Read 2 more answers
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an
sleet_krkn [62]

Answer:

-6re−r [sin(6θ) - cos(6θ)]

Step-by-step explanation:

the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

x = e−r sin(6θ), y = er cos(6θ)

δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)

δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)

∂(x, y) /∂(r, θ) =  δx/δθ × δy/δr - δx/δr × δy/δθ

= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]

= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))

= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]

= -6re−r [sin(6θ) - cos(6θ)]     since  [cos²(6θ) + sin²(6θ)] = 1

6 0
3 years ago
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