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3 years ago

Add (4x^2-xy-y^2) and (x^2+5xy+8y^2) simplify your answer:

Mathematics

2 answers:

bazaltina [42]3 years ago

8 0

Answer:

To find a, b, and c, rewrite in the standard form ax2+bx+c=0ax2+bx+c=0.

a=1, b=3, c=0

sukhopar [10]3 years ago

4 0

Answer:

4x^2-4xy-3\frac{y^2}{2}x^2-5xy-3y^2 is 2

−3x^2 y^2 +8x^2 −18xy−6y^2

Step-by-step explanation:

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Sveta_85 [38]

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3/14

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3 years ago

Please Help!!

Genrish500 [490]

Given

$a\sqrt{x+b}+c=d$

we have

$\sqrt{x+b}=\dfrac{d-c}{a}$

Squaring both sides, we have

$x+b=\dfrac{(d-c)^2}{a^2}$

And finally

$x=\dfrac{(d-c)^2}{a^2}-b$

Note that, when we square both sides, we have to assume that

$\dfrac{d-c}{a}>0$

because we're assuming that this fraction equals a square root, which is positive.

So, if that fraction is positive you'll actually have roots: choose

$a=1,\ b=0,\ c=2,\ d=6$

and you'll have

$\sqrt{x}+2=6 \iff \sqrt{x}=4 \iff x=16$

Which is a valid solution. If, instead, the fraction is negative, you'll have extraneous roots: choose

$a=1,\ b=0,\ c=10,\ d=4$

and you'll have

$\sqrt{x}+10=4 \iff \sqrt{x}=-6$

Squaring both sides (and here's the mistake!!) you'd have

$x=36$

which is not a solution for the equation, if we plug it in we have

$\sqrt{x}+10=4 \implies \sqrt{36}+10=4 \implies 6+10=4$

Which is clearly false

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3 years ago

What equation represents the line that passes through the points(6,-3) and (-4,-9)?

marysya [2.9K]

M = - 3 + 9 / 6 + 4
m = 6 / 10
m = 3/5

- 3 = 3/5 ( 6 ) + c
- 15 = 18 + 5c
5c = - 33
c = - 33/5
y = 3/5x - 33/5

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3 years ago

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Hitman42 [59]

Answer:

A. cd4

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horsena [70]

Answer:

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3 years ago

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