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3 years ago

Put the equation in standard form with a positive X coefficient y + 3 = 3(x+5)

Mathematics

2 answers:

Tems11 [23]3 years ago

7 0

Answer:

3x + y = -12.

Step-by-step explanation:

y + 3 = 3(x+5)

y + 3 = 3x + 15

3x + 15 = y + 3

3x - y + 15 = 3

3x + y = -12

Readme [11.4K]3 years ago

3 0

Answer:

3x - y = - 12

Step-by-step explanation:

The equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

given

y + 3 = 3(x + 5) ← distribute

y + 3 = 3x + 15 ( subtract y from both sides )

3 = 3x - y + 15 ( subtract 15 from both sides )

- 12 = 3x - y, thus

3x - y = - 12 ← in standard form

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3 years ago

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3 years ago

Read 2 more answers

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elena55 [62]

$45.50
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3 years ago

See attachment to help.

GenaCL600 [577]

B.
because {1,1,3,5,8}
{2,2,4,6,9}

you're just counting the numbers after the one that was given to you.

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3 years ago

How do i solve for x? *you have to show work*

eduard

1) Factor the equation first:

$2e^{2x} + 4xe^{2x} = 0$
$e^{2x}(2 + 4x) = 0$

So either $e^{2x} = 0$ or $2 + 4x = 0$ . But $e^{2x} = 0$ can never equal 0! So, we only need to solve for the second equation:

$2 + 4x = 0$
$\bf x = {-\frac{1}{2}}$

2) First, single out the term with the exponent:

$2 + 3(4^{2x - 1}) = 43$
$3(4^{2x - 1}) = 41$
$4^{2x - 1} = \frac{41}{3}$

Now, taking the log base 4 of each side gives us

$2x - 1 = \log_4 \frac{41}{3}$

and we can now solve for x:

$2x - 1 = \log_4 \frac{41}{3}$
$2x = 1 + \log_4 \frac{41}{3}$
$2x = \log_4 4 + \log_4 \frac{41}{3}$
$2x = \log_4 \frac{164}{3}$
$x = \frac{1}{2} \log_4 \frac{164}{3}$
$x = \log_4 \sqrt \frac{164}{3}$
$\bf x = \log_4 2 \sqrt \frac{41}{3}$

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3 years ago

Put the equation in standard form with a positive X coefficient y + 3 = 3(x+5) ​

Put the equation in standard form with a positive X coefficient y + 3 = 3(x+5)