Which function of x has the least value for the y-intercept?

Which expression is equivalent to logc x^2-1/5x?

fenix001 [56]

Answer:

log(x²-1)-(log5+logx)

Step-by-step explanation:

For this problem we need to make use of two existing logarithmic properties. These are:

logM/N is equal to logM - logN

logMN is equal to logM+logN

Following the first property, we can simplify the expression to:

log(x²-1)-log5x

Then, we will use the second property to the term 5x. The final form of the expression would then be:

log(x²-1)-(log5+logx)

Thus the answer is log(x²-1)-(log5+logx)....

4 0

3 years ago

Jelly makes $9 per hour. she had worked 8 hours so far this week. How much has she made so far this week? Write an algebraic exp

Brrunno [24]

Answer:

8×9=x x=72 * don't forget Mark me as Brainliest

3 0

3 years ago

HELP ILL GIVE 10 PTS quick pleaseee

Sladkaya [172]

Answer: I’m pretty sure it’s the 4th one.

Step-by-step explanation: because 1 and 2 have infinity solutions but 3 has no solution 4 comes up with x=-11

Can you give brainiest? :)

6 0

3 years ago

Please solve this please

ale4655 [162]

Answer:

C) $\frac{2z+15}{6x-12y}$

E) $\frac{7d+5}{15d^2+14d+3}$

F) $\frac{-7a-b}{6b-4a}$

Step-by-step explanation:

C)

One is given the following equation

$\frac{z+1}{x-2y}-\frac{2z-3}{2x-4y}+\frac{z}{3x-6y}$

In order to simplify fractions, one must convert the fractions to a common denominator. The common denominator is the least common multiple between the given denominators. Please note that the denominator is the number under the fraction bar of a fraction. In this case, the least common multiple of the denominators is ( $6x-12y$ ). Multiply the numerator and denominator of each fraction by the respective value in order to convert the fraction's denominator to the least common multiple,

$\frac{z+1}{x-2y}-\frac{2z-3}{2x-4y}+\frac{z}{3x-6y}$

$\frac{z+1}{x-2y}*\frac{6}{6}-\frac{2z-3}{2x-4y}*\frac{3}{3}+\frac{z}{3x-6y}*\frac{2}{2}$

Simplify,

$\frac{z+1}{x-2y}*\frac{6}{6}-\frac{2z-3}{2x-4y}*\frac{3}{3}+\frac{z}{3x-6y}*\frac{2}{2}$

$\frac{6z+6}{6x-12y}-\frac{6z-9}{6x-12y}+\frac{2z}{6x-12y}$

$\frac{(6z+6)-(6z-9)+(2z)}{6x-12y}$

$\frac{6z+6-6z+9+2z}{6x-12y}$

$\frac{2z+15}{6x-12y}$

E)

In this case, one is given the problem that is as follows:

$\frac{2}{3d+1}-\frac{1}{5d+3}$

Use a similar strategy to solve this problem as used in part (c). Please note that in this case, the least common multiple of the two denominators is the product of the two denominators. In other words, the following value: ( $(3d+1)(5d+3)$ )

$\frac{2}{3d+1}-\frac{1}{5d+3}$

$\frac{2}{3d+1}*\frac{5d+3}{5d+3}-\frac{1}{5d+3}*\frac{3d+1}{3d+1}$

Simplify,

$\frac{2}{3d+1}*\frac{5d+3}{5d+3}-\frac{1}{5d+3}*\frac{3d+1}{3d+1}$

$\frac{2(5d+3)}{(3d+1)(5d+3)}-\frac{1(3d+1)}{(5d+3)(3d+1)}$

$\frac{10d+6}{(3d+1)(5d+3)}-\frac{3d+1}{(5d+3)(3d+1)}$

$\frac{(10d+6)-(3d+1)}{(3d+1)(5d+3)}$

$\frac{10d+6-3d-1}{(3d+1)(5d+3)}$

$\frac{7d+5}{(3d+1)(5d+3)}$

$\frac{7d+5}{15d^2+14d+3}$

F)

The final problem one is given is the following:

$\frac{3a}{2a-3b}-\frac{a+b}{6b-4a}$

For this problem, one can use the same strategy to solve it as used in parts (c) and (e). The least common multiple of the two denominators is ( $6b-4a$ ). Multiply the first fraction by a certain value to attain this denomaintor,