3 years ago

Is the problem I showed congruent

Mathematics

1 answer:

dimulka [17.4K]3 years ago

7 0

Answer:

it is very congruent of you k

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sp2606 [1]

$\displaystyle\\Answer:\ x=\sqrt{2}\ \ \ \ \ x=-\sqrt{2}\ \ \ \ \ x=\frac{ln5}{2}$

Step-by-step explanation:

$5x^2-x^2e^{2x}+2e^{2x}=10$

$Let:\ x^2=t \ \ \ \ \ e^{2x}=v\\Hence,\\5t-tv+2v^2=10\\5t-tv+2v^2-10=10-10\\5t-tv-10+2v^2=0\\t(5-v)-2(5-v)=0\\(5-v)(t-2)=0\\5-v=0\\5-v+v=0+v\\5=v\\5=e^{2x}$

Prologarithmize both parts of the equation:

$ln5=lne^{2x}\\ln5=2x*lne\\ln5=2x*1\\ln5=2x\\$

Divide both parts of the equation by 2:

$\displaystyle\\x=\frac{ln5}{2}$

$t-2=0\\t-2+2=0+2\\t=2\\x^2=2\\x=\sqrt{2} \\x=-\sqrt{2}$

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1 year ago

If NTM is congruent to OUM, SP is congruent to SQ, S is the center of the circle, and OM = 18, find NT. Round the answer to the

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OM=18, so OQ=QM=18/2=9.
Given QU=8
from figure OQU is a right angled triangle , so OU^2=OQ^2 + QU^2
OU^2 = 9*9 + 8*8 = 81+72=153;
OU=sqrt(153) = 12.37 =13(approx);
From given statements of congruent NT and OU will also be congruent or identical. So, NT=OU=13

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