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irina1246 [14]
3 years ago
13

Solve.

Mathematics
1 answer:
Eva8 [605]3 years ago
4 0

2x^2+x-1=2\ \ \ |-2\\\\2x^2+x-3=0\\\\2x^2+3x-2x-3=0\\\\x(2x+3)-1(2x+3)=0\\\\(2x+3)(x-1)=0\iff 2x+3=0\ \vee\ x-1=0\\\\2x+3=0\ \ |-3\\2x=-3\ \ \ |:2\\x=-\frac{3}{2}\\\\x-1=0\ \ \ |+1\\x=1

Answer: C 1, -3/2

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169= (24/2)^2 +(d/2)^2
OlgaM077 [116]

Answer: d = ±10

<u>Step-by-step explanation:</u>

169 = (\frac{24}{2})^{2} + (\frac{d}{2})^{2}

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169 = 144  + \frac{d^{2}}{4}

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3 years ago
What are the coordinates of R’, the image of R(-4,3) after a reflection in the line y=x?
navik [9.2K]

Answer:

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Step-by-step explanation:

we know that

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we have

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3 years ago
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WITCHER [35]
Write the coeeficientes of the polynomial in order:

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------------------------

After some trials you probe with 5

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-----------------------------
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Given that the residue is 0, 5 is a root.

The quotient is x^2 + 6 = 0, which does not have a real root.

Therefore, 5 is the only root. You can prove it by solving the polynomial x^2 + 6 = 0.
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Step-by-step explanation:

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