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Anit [1.1K]
3 years ago
12

If BCDE is congruent to OPQR then DE is congruent to what?

Mathematics
1 answer:
irga5000 [103]3 years ago
5 0

Answer:

QR

Step-by-step explanation:

Because if BC is congruent with OP then DE will congruent with QR

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Answer: I don't know what you meant by remainder but i hope this helps :)

x=\frac{-3+\sqrt{29}}{4},\:x=-\frac{3+\sqrt{29}}{4}\\

Step-by-step explanation:

\left(4x^2+7x-1\right)=\left(4+x\right)\\\mathrm{Refine}\\4x^2+7x-1=4+x\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\4x^2+7x-1-x=4+x-x\\Simplify\\4x^2+6x-1=4\\\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}\\4x^2+6x-1-4=4-4\\\mathrm{Simplify}\\4x^2+6x-5=0\\\mathrm{For\:}\quad a=4,\:b=6,\:c=-5:\\\quad x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}

\frac{-6+\sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}\\=\frac{-6+\sqrt{6^2+4\times \:4\times \:5}}{2\times \:4}\\=\frac{-6+\sqrt{116}}{2\times \:4}\\=\frac{-6+\sqrt{116}}{8}\\\\Let\: simplify\: ; -6+2\sqrt{29}\\=-2\times \:3+2\sqrt{29}\\=2\left(-3+\sqrt{29}\right)\\=\frac{2\left(-3+\sqrt{29}\right)}{8}\\=\frac{-3+\sqrt{29}}{4}\\

\frac{-6-\sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}\\\\=\frac{-6-\sqrt{6^2+4\times \:4\times \:5}}{2\times \:4}\\\\=\frac{-6-\sqrt{116}}{2\times \:4}\\\\=\frac{-6-2\sqrt{29}}{8}\\\\=-\frac{2\left(3+\sqrt{29}\right)}{8}\\\\=-\frac{3+\sqrt{29}}{4}\\\\\\x=\frac{-3+\sqrt{29}}{4},\:x=-\frac{3+\sqrt{29}}{4}

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