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ad-work [718]
3 years ago
6

2x+7y=1 X+5y=2 answer using elimination method

Mathematics
1 answer:
miskamm [114]3 years ago
5 0
\left \{ {{2x+7y=1} \atop {x+5y=2\ \ |*(-2)}} \right. \\\\
 \left \{ {{2x+7y=1} \atop {-2x-10y=-4}} \right. \\+----\\elimination\ method\\\\
-3y=-3\ \ \ | divide\ by\ -3\\
y=1\\\\
x=2-5y=2-5=-3\\\\
solution:\\ \left \{ {{y=1} \atop {x=-3}} \right.
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A pair of pants on sale for 25% off if they're original price was $40 what is the new sale price
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Transform x2 + 6x + 8 = 0 into the form (x − p)2 = q?
liraira [26]
So hmmm x²+6x+8=0

alrite.. let's do some grouping now
( x² + 6x + [?]²) + 8 = 0

notice above, we have a missing fellow in order to get a perfect square trinomial... hmm who would that be?

let's take a peek at the middle guy of the trinomial.. 6x.. hmmm let's factor it, 2*3*x, wait a minute!  2 * 3 * x... we already have x² on the left-side, since the middle term is just 2 * the square root of the other two terms, that means that the guy on the right, our missing guy must be "3"

alrite, let's add 3² then, however, bear in mind that, all we're doing is borrowing from our very good friend Mr Zero, 0

so if we add 3², we also have to subtract 3², let's do so

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(x+3)²=3² - 8

(x+3)² = 1
5 0
3 years ago
Read 2 more answers
I NEED HELP PLEASE, THANKS! :)
Elena L [17]

Answer:

First option is the right choice.

Step-by-step explanation:

\frac{1}{\cos \left(x\right)+1}+\frac{1}{\cos \left(x\right)-1}\\\\\frac{\cos \left(x\right)-1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}+\frac{\cos \left(x\right)+1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}\\\\=\frac{\cos \left(x\right)-1+\cos \left(x\right)+1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}\\\\=\frac{2cos\left(x\right)}{cos^2\left(x\right)-1}\\\\=\frac{2cos\left(x\right)}{sin^2\left(x\right)}

=2\cdot \frac{cos\left(x\right)}{sin\left(x\right)}\cdot \frac{1}{sin\left(x\right)}\\\\=-2cot(x)csc(x)

Best Regards!

7 0
3 years ago
Recall the equation for a circle with center ( h , k ) (h,k) and radius r r. At what point in the first quadrant does the line w
garri49 [273]
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Answer:

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Step-by-step explanation:

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