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3 years ago

So i don't understand how factoring equations work in algebra 1

Mathematics

2 answers:

telo118 [61]3 years ago

7 0

Answer:

this will help https://www.google.com/search?q=factoring+equations&oq=factoring+equations&aqs=chrome..69i57j0l6j69i61.464j0j7&sourceid=chrome&ie=UTF-8#kpvalbx=_w9Z8XvnVCsq7tgXx3SM51

Step-by-step explanation:

its in the video https://www.google.com/search?q=factoring+equations&oq=factoring+equations&aqs=chrome..69i57j0l6j69i61.464j0j7&sourceid=chrome&ie=UTF-8#kpvalbx=_w9Z8XvnVCsq7tgXx3SM51

sdas [7]3 years ago

6 0

It’s basically like the opposite of simplifying an equation.

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Draw a diagram for this fraction<br><br>2 2/3

LiRa [457]

Answer:

OOO

Step-by-step explanation:

Where OOO is == 1.

OO is 2/3 of OOO.

7 0

2 years ago

Answer please GIVING BRAINLIEST

Kaylis [27]

Answer:

see below

Step-by-step explanation:

defining variables:

since you are trying to find t shirt cost and jeans cost, your variables would be

the cost of the t shirt and the cost of jeans, denoted by any variable you want.

I'll use x and y since they are the most common variables (x = shirt y = jeans)

Equation

4 shirts + 5 jeans = 160 dollars

4x + 5y = 160 (the cost of 4 shirts + the cost of 5 jeans is $160)

the cost of 2 shirts + 6 jeans = $150

2x + 6y = 150

3 0

3 years ago

Read 2 more answers

You start at (6, 7). You move up 2 units. Where do you end?

Vikki [24]

Answer:

(6,9) nice

Step-by-step explanation:

8 0

3 years ago

Hey Mahfia, please help! Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at

Dafna1 [17]

Answer:

$\huge\boxed{ \red{ \boxed{ \tt{ \frac{ {y}^{2} }{ {10}^{2} } - \frac{ {x}^{2} }{ {12}^{2} } = 1}}}}$

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

conic sections
PEMDAS

<h3>tips and formulas:</h3>

$\sf hyperbola \:equation : \\ \sf \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1$
vertices of hyperbola:(±a,0) and (0,±b) if reversed
$\sf \: asymptotes : \\ y = \pm\frac{b}{a} x$

<h3>given:</h3>

vertices: (0,±10)
the hyperbola equation is inversed since the vertices is (0,±10)
asymptotes: $\pm \frac{5}{6}x$

<h3>let's solve:</h3>

the asymptotes are in simplest and we know b is ±10

according to the question

$y = \sf \frac{5 \times 2}{6 \times 2} x \\ y = \frac{10}{12} x$

therefore we got

a=12
b=10

note: the equation will be inversed

let's create the equation:

$\sf substitute \: the \: value \: of \: a \: and \: b : \\ \sf \frac{ {y}^{2} }{ {10}^{2} } - \frac{ {x}^{2} }{ {12}^{2} } = 1$

3 0

3 years ago

What is the equation of a line that passes through the point (6, 1) and is perpendicular to the line whose equation is y=−2x−6y=

Mandarinka [93]

Lets get the negative reciprocal of the slope so we get
$m = \frac{1}{2}$
Now the equation that is perpendicular is
$y = \frac{1}{2} x + b$
Lets solve for b by plugging in the point (x,y)=(6,1)
$1 = \frac{1}{2} (6) + b \\ 1 = 3 + b \\ 1 - 3 = b \\ b = - 2$
Now plug that into the equation. So the equation of the line that is perpendicular and
passes through the point (6, 1) is
$y = \frac{1}{2} x - 2$

6 0

3 years ago