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3 years ago

Describe how to graph a line given an equation in standard form.

Mathematics

1 answer:

Archy [21]3 years ago

3 0

This is pretty simple. So a standard form equation is probably something like this: 2x+8y=10.
We have to get that equation right there into a slope intercept, which is just isolating the variable y.
2x+8y=16
-2x -2x Subtract the x value.
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(8y=16-2x)/8 Divide the whole equation by 8 to isolate the y value.
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y=2-1/4x
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The fraction with the variable x is the slope. In this case, the slope is -1/4, so meaning it goes down one unit and goes right 4.
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The 2 is the y-intercept. (0, 2)

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BabaBlast [244]

$\underline{\bf{Given \:equation:-}}$

$\\ \sf{:}\dashrightarrow ax^2+by+c=0$

$\sf Let\:roots\;of\:the\: equation\:be\:\alpha\:and\beta.$

$\sf We\:know,$

$\boxed{\sf sum\:of\:roots=\alpha+\beta=\dfrac{-b}{a}}$

$\boxed{\sf Product\:of\:roots=\alpha\beta=\dfrac{c}{a}}$

$\underline{\large{\bf Identities\:used:-}}$

$\boxed{\sf (a+b)^2=a^2+2ab+b^2}$

$\boxed{\sf (√a)^2=a}$

$\boxed{\sf \sqrt{a}\sqrt{b}=\sqrt{ab}}$

$\boxed{\sf \sqrt{\sqrt{a}}=a}$

$\underline{\bf Final\: Solution:-}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}$

$\bull\sf Apply\: Squares$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2= (\sqrt{\alpha})^2+2\sqrt{\alpha}\sqrt{\beta}+(\sqrt{\beta})^2$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2 \alpha+\beta+2\sqrt{\alpha\beta}$

$\bull\sf Put\:values$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2=\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}}$

$\bull\sf Simplify$

$\\ \sf{:}\dashrightarrow \underline{\boxed{\bf {\sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\sqrt{\dfrac{-b}{a}}+\sqrt{2}\dfrac{c}{a}}}}$

$\underline{\bf More\: simplification:-}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{-b}}{\sqrt{a}}+\dfrac{c\sqrt{2}}{a}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{a}\sqrt{-b}+c\sqrt{2}}{a}$

$\underline{\Large{\bf Simplified\: Answer:-}}$

$\\ \sf{:}\dashrightarrow\underline{\boxed{\bf{ \sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\dfrac{\sqrt{-ab}+c\sqrt{2}}{a}}}}$

5 0

2 years ago

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Name The postulate or theorem you can use to prove the triangles congruent

Zielflug [23.3K]

The answer is HL. HL is a theorem stating that if the hypotenuse and one leg are congruent to another pair of hypotenuse and leg then the triangles themselves are congruent. Hope this helps.

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3 years ago

All of the following are equivalent except _____.

Inessa05 [86]

(5_5)x is not equalant

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2 years ago

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Find the Prime Factorization of -165

Ghella [55]

Answer:

(-1)(11)(3)(5)

Step-by-step explanation:

First off: -165 separates into the factors (-1) and (165). Next, we divide 165 by 15, obtaining 15(11).

So far, we have (-1)(11)(15).

15 is still factorable. We get:

(-1)(11)(3)(5), which is the desired prime factorization of -165.

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3 years ago

Four-fiths of a number minus two is greater then three-tenths of twice of that number

pashok25 [27]

Answer:

Step-by-step explanation:

4/5 x - 2 > 3/10 * 2x

4/5 x - 2 > 3/5 x

4/5 x - 2 - 3/5x> 0

1/5x > 2

x > 10

7 0

2 years ago