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3 years ago

Which of the following are solutions to the equation below?

Mathematics

2 answers:

snow_lady [41]3 years ago

5 0

Answer:

Step-by-step explanation:

$5x^2-2x+16=4x^2+6x\qquad\text{subtract}\ 4x^2\ \text{from both sides}\\\\x^2-2x+16=6x\qquad\text{subtract}\ 6x\ \text{from both sides}\\\\x^2-8x+16=0\\\\\text{Put the values of}\ x\ \text{to the equation and check the equality:}\\\\A.\ x=-6\\\\(-6)^2-8(-6)+16=36+48+16=100\neq0\\\\B.\ x=6\\\\6^2-8(6)+16=36-48+16=4\neq0\\\\C.\ x=-3\\\\(-3)^2-8(-3)+16=9+24+16=49\neq0\\\\D.\ x=-4\\\\(-4)^2-8(-4)+16=16+32+16=64\neq0\\\\E.\ x=4\\\\4^2-8(4)+16=16-32+16=0\\\\F.\ x=18\\\\18^2-8(16)+16=324-128+16=212\neq0$

Leona [35]3 years ago

3 0

Answer:

Step-by-step explanation:

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What the answer now to the question

Lostsunrise [7]

Step-by-step explanation:

Hello!!!

Given, VXW is a Right angled triangle where WV =4 and XV =5.

now, by taking reference angle as angle W, and using tangent we get;

p=5

b=4

again, 

tan thita =p/b

tan thita = 5/4

so, tan thita =1.25.

now, to find angle W, we should do tan thita inverse;

or, (tan thita)-1= tan thita-1(1.25)

or, (tan thita) -1 =51.3401°.

Now, by rounding off we get,

tan thita (angle W )= 51°.

Hope it helps....

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Read 2 more answers

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so
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