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liq [111]
2 years ago
12

Evaluate 3x2 - 4 when χ= 2. Α. 2 Ο Ο Ο Ο Β. 8 C. 12 Ο D. 32

Mathematics
1 answer:
solniwko [45]2 years ago
3 0

Answer:

D. 32.

Step-by-step explanation:

3x^2 - 4\\= 3^2 * 2\\= 6^2 - 4\\= 36 - 4\\= 32.

hope this helps.

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$34 jacket 10% markup
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The value of
marusya05 [52]

\large\underline{\sf{Solution-}}

We have to find out the value of the fraction.

<u>Let us assume that:</u>

\sf \longmapsto x =2 +   \dfrac{1}{2 +  \dfrac{1}{2 +  \dfrac{1}{2 + ... \infty} } }

<u>We can also write it as:</u>

\sf \longmapsto x =2 + \dfrac{1}{x}

\sf \longmapsto x =\dfrac{2x + 1}{x}

\sf \longmapsto  {x}^{2}  =2x + 1

\sf \longmapsto {x}^{2}  - 2x - 1 = 0

<u>Comparing </u>the given <u>equation</u> with <u>ax² + bx + c = 0,</u> we get:

\sf \longmapsto\begin{cases} \sf a =1 \\ \sf b =  - 2 \\ \sf c =  - 1 \end{cases}

<u>By quadratic formula:</u>

\sf \longmapsto x =  \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}

\sf \longmapsto x =  \dfrac{2 \pm \sqrt{ {( - 2)}^{2} - 4(1)( - 1)} }{2 \times 1}

\sf \longmapsto x =  \dfrac{2 \pm \sqrt{4 + 4} }{2 \times 1}

\sf \longmapsto x =  \dfrac{2 \pm \sqrt{8} }{2}

\sf \longmapsto x =  \dfrac{2 \pm2 \sqrt{2} }{2}

\sf \longmapsto x = 1 \pm\sqrt{2}

\sf \longmapsto x = \begin{cases} \sf 1  + \sqrt{2} \\ \sf 1 -  \sqrt{2}  \end{cases}

<u>But </u><u>"</u><u>x"</u><u> cannot be negative. Therefore:</u>

\sf :\implies x = 1 + \sqrt{2}

So, the value of the fraction is 1 + √2.

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Which of the following reasons can be used to justify statement #4 in the proof?
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Answer:

D.

Step-by-step explanation:

CPCTC which stands for "corresponding parts of congruent triangles are congruent". Since ΔMQN and ΔPQN have already been proved congruent, So every angle and sides of these two triangle corresponding with each other would be congruent.

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The word geometry comes from the greek words geo, which means "earth," and metron, which means "a measuring of." given this info
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The geometry was developed by a Greek mathematician called Euclid and developed to meet some practical need in surveying, construction, astronomy, and various crafts.

In this question,

Early geometry was a collection of empirically discovered principles concerning lengths, angles, areas, and volumes, which were developed to meet some practical need in surveying, construction, astronomy, and various crafts.

It was developed by a Greek mathematician called Euclid. This branch of geometry deals with terms like points, lines, surfaces, dimensions of the solids, etc.,

In several ancient cultures there developed a form of geometry suited to the relationships between lengths, areas, and volumes of physical objects. This geometry was codified in Euclid's Elements about 300 bce on the basis of 10 axioms, or postulates, from which several hundred theorems were proved by deductive logic.

Hence we can conclude that the geometry was developed by a Greek mathematician called Euclid and developed to meet some practical need in surveying, construction, astronomy, and various crafts.

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brainly.com/question/28247330

#SPJ4

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