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3 years ago

8

Determine whether the given point (2, -2) lies on the line 3x – y = 4.

1 answer:

AVprozaik [17]3 years ago

8 0

Answer:

Step-by-step explanation:

3(2) + 2 = 4

6 + 2 = 4

8 is not equal to 4

Therefore (2,-2) does not equal

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7x - 20 = 2x - 3(3x + 2)

kirill [66]

Answer:

7x-20=2x-3(3x+2)

We move all terms to the left:

7x-20-(2x-3(3x+2))=0

We calculate terms in parentheses: -(2x-3(3x+2)), so:

2x-3(3x+2)

We multiply parentheses

2x-9x-6

We add all the numbers together, and all the variables

-7x-6

Back to the equation:

-(-7x-6)

We get rid of parentheses

7x+7x+6-20=0

We add all the numbers together, and all the variables

14x-14=0

We move all terms containing x to the left, all other terms to the right

14x=14

x=14/14

x=1

Step-by-step explanation:

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almond37 [142]

Answer:

Option D, $x > 6$

Step-by-step explanation:

<u>Step 1: Subtract 8 from both sides </u>

<u /> $x + 8 > 14$

$x + 8 - 8 > 14 - 8$

$x > 6$

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Answer: it the area of it

Step-by-step explanation:

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ln(y+1)-lny=1+3lnx . Express y in terms of x, in a form not involving logarithms. (4 marks, Maths CIE A level)

Delicious77 [7]

$remember$
$ln(x)=log_e(x)$
$and$
$log_x(a)-log_x(b)= log_x( \frac{a}{b} )$
$and$
$log_a(b)=c$ means $a^c=b$
$and$
$blog(a)=log(a^b)$

$so$

$ln(y+1)-ln(y)=1+3ln(x)$
$ln( \frac{y+1}{y} )=1+ln(x^3)$

$minus$ $ln(x^3)$ $from$ $both$ $sides$

$ln( \frac{y+1}{y}-ln(x^3) )=1$
$ln( \frac{y+1}{yx^3})=1$
$log_e( \frac{y+1}{yx^3})=1$

$translate$

$e^1=\frac{y+1}{yx^3}$
$e=\frac{y+1}{yx^3}$

$times$ $both$ $sides$ $by$ $yx^3$

$eyx^3=y+1$

$minus$ $y$ $from$ $both$ $sides$

$eyx^3-y=1$
$y(ex^3-1)=1$

$divide$ $both$ $sides$ $by$ $(ex³-1)$

$y= \frac{1}{ex^3-1}$

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