We determine line m as follows:
*First, by theorem we have the following:

Here m1 & m2 are the slopes of two perpendicular lines. For all lines that are perpendicular that is true, so we calculate the slope of line m using the slope of the function given [Which has a slope of 7/4]:

So, the slope of line m is -4/7. Now, using this slope and the point (-1, 4) we replace in the following expression:

Here x1, y1 & m1 are the x-component of the point, the y-component of the point, and the slope of the line respectively, so we replace and solve for y:


And that last function of y is the line m.
Find the common denominator of 6 and 4.
6: 6, 12...
4: 8, 12...
We can use 12 for this example.
5/6 to ?/12.
12 / 6 = 2, therefore do 5 x 2.
5 x 2 = 10. 10/12.
1/4 to ?/12.
12 / 4 = 3, therefore do 3 x 1.
3 x 1 = 3. 3/12.
13/12. 1 1/12.
I believe you ran out of space or didn't put answer D. I researched this question and concluded that the other choice is 1 and 1/12.
The answer is 1 and 1/12.
They would cost $8.75. If you take out 30% from the original price, that is how much they would be.
Answer:
x = 2
Step-by-step explanation:
(x² logₓ27) log₉x = x + 4
Use change of base formula.
x² (log 27 / log x) (log x / log 9) = x + 4
Simplify.
x² (log 27 / log 9) = x + 4
x² (log 3³ / log 3²) = x + 4
x² (3 log 3 / (2 log 3)) = x + 4
x² (3 / 2) = x + 4
3x² = 2x + 8
3x² − 2x − 8 = 0
(x − 2) (3x + 4) = 0
x = 2 or -4/3
Since x > 0, x = 2.