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3 years ago

Which symbol will make |-6| ? -6 true? A > B < C =

Mathematics

1 answer:

LiRa [457]3 years ago

3 0

A because | | <- around a number means the absolute value which in this case is 6.

|-6| > -6
6 > -6

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Completing the sqaure x^2+14x+34

DiKsa [7]

Answer:

-7-√15

Step-by-step explanation:

First move the constant to the right hand side and change it's sign

x^2+14x=-34

Then add (14/2)^2 or 7^2 to both sides

x^2+14x+49=-34+49

Then factor the expression

(x+7)^2=-34+49

Then solve (x+7)^2=15

You should get:

x=-√15-7

x=√15-7

7 0

3 years ago

Use the properties of logarithms to expand the expression as a sum or difference, and/or constant multiple of logarithms. (Assum

Alecsey [184]

Answer:

$\log_5(9)+\frac{1}{2}\log_5(x)-3\log_5(y)$

Step-by-step explanation:

$\log_5(\frac{9\sqrt{x}}{y^3})$

First rule I'm going to use is the quotient rule:

$\log_b(\frac{m}{n})=\log_b(m)-\log_b(n)$

$\log_5(9\sqrt{x})-\log_5(y^3)$

Secondly, I'm going to rewrite the radical.

$\sqrt{x}=x^\frac{1}{2}$

$\log_5(9x^\frac{1}{2})-\log_5(y^3)$

Third, I'm going to use the product rule on the first term:

$\log_b(mn)=\log_b(m)+\log_b(n)$

$\log_5(9)+\log_5(x^\frac{1}{2})-\log_5(y^3)$

Fourth, I'm going to use power rule for both of the last two terms:

$\log_b(m^r)=r\log_b(m)$

$\log_5(9)+\frac{1}{2}\log_5(x)-3\log_5(y)$

6 0

2 years ago

Find the slope of the line passing through the points (7,-8) and (1, -1).

natita [175]

Answer:

5= - 7/6

Hope this helps

6 0

2 years ago

Read 2 more answers

Change 108% to a mixed number and reduce, if possible.

monitta

27/25 is in the lowest form

5 0

3 years ago

Read 2 more answers

WHAT IS X³-27 SIMPLIFIED

Eduardwww [97]

Answer:

It is (x - 3)³ - 9x(3 - x)

Step-by-step explanation:

Express 27 in terms of cubes, 27 = 3³:

$= {x}^{3} - {3}^{3}$

From trinomial expansion:

${(x - y)}^{3} = (x - y)(x - y)(x - y) \\$

open first two brackets to get a quadratic equation:

${(x - y)}^{3} = ( {x}^{2} - 2xy + {y}^{2} )(x - y)$

expand further:

${(x - y)}^{3} = {x}^{3} - y {x}^{2} - 2y {x}^{2} + 2x {y}^{2} + x {y}^{2} - {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - {y}^{3} + 3x {y}^{2} - 3y {x}^{2} \\ {(x - y)}^{3} = {x}^{3} - {y}^{3} + 3xy(y - x) \\ \\ { \boxed{( {x}^{3} - {y}^{3} ) = {(x - y)}^{3} - 3xy(y - x)}}$

take y to be 3, then substitute:

$( {x}^{3} - 3^3) = {(x - 3)}^{3} - 9x(3 - x)$

5 0

2 years ago