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Ronch [10]
3 years ago
8

Which of the given sets of vectors in r3, with the standard inner product, are orthogonal, orthonormal, or neither? (choose the

option that best describes the set.) (a) 1 2 0 1 2 , − 1 2 0 1 2 , 0 1 0 orthogonal orthonormal neither?
Mathematics
1 answer:
Nonamiya [84]3 years ago
4 0
Jjjjjjjjjjjjjjjjjjjjjj
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allsm [11]

I believe the answer is 1064.

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I need help asap pls and thank you ;)
olga55 [171]

Answer:

\text{Length of AB is }\frac{ah}{a+h}

Step-by-step explanation:

Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.

we have to find the length of AB.

Let the side of square i.e AB is x units.

As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°

⇒ CP||MP||AB

In ΔMNP and ΔCND

∠NCD=∠NMP     (∵ corresponding angles)

∠NDC=∠NPM     (∵ corresponding angles)

By AA similarity rule,  ΔMNP~ΔCND

Also, ΔKAP~ΔKPM by similarity rule as above.

Hence, corresponding sides are in proportion

\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\

KA(\frac{h}{x}-1)=AP

ND(\frac{h}{x}-1)=PD

Adding above two, we get

(KA+ND)(\frac{h}{x}-1)=(AP+PD)

⇒ (KN-AD)=\frac{x}{(\frac{h}{x}-1)}

⇒ a-x=\frac{x}{(\frac{h}{x}-1)}

⇒ a-x=\frac{x^2}{h-x}

⇒ x^2=ah-ax-xh+x^2

⇒ x(h+a)=ah

⇒ x=\frac{ah}{a+h}

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3 years ago
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Zigmanuir [339]

Answer:

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Is the ordered pair (-2,-2) a solution to th fiction y= 3x-8
Tema [17]

Answer:-24

Step-by-step explanation:

You multiply your numbers without the - or the +

You check ehich number is bigger and you use it symbol for the answer

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