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damaskus [11]
2 years ago
15

Solve this quadratic inequalitie with detailed full explanation

Mathematics
2 answers:
ANTONII [103]2 years ago
7 0

x^2\leq16\\x\leq 4 \wedge x\geq -4\\x\in\langle-4,4\rangle

different approach:

x^2\leq16\\x^2-16\leq0\\(x-4)(x+4)\leq0

see attachment

x\in\langle-4,4\rangle

Blizzard [7]2 years ago
5 0

~~~~~~x^2 \leq 16\\\\\implies -\sqrt{16} \leq x\leq  \sqrt{16}\\\\\implies -4\leq x \leq 4\\\\\text{Interval notation:}~ [-4,4]

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I need help with this because I'm not very good at understanding this.
HACTEHA [7]

Answer:

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Step-by-step explanation:

"y=mx+b" is the equation

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brilliants [131]
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2x+1=73.
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</span>
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6 0
3 years ago
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Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.
lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ

Likewise:

\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ

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Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

m\angle PAC=m\angle PAB+m\angle BAC

Likewise:

m\angle QAB=m\angle QAC+m\angle BAC

Since ∠QAC ≅ ∠PAB:

m\angle PAC=m\angle QAC+m\angle BAC

And by substitution:

m\angle PAC=m\angle QAB

Thus:

\angle PAC\cong \angle QAB

Then by SAS Congruence:

\Delta PAC\cong \Delta BAQ

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PC\cong BQ

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3 years ago
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