Plz help me solve this question algebraic expression 64p^4+81q^4

Mathematics

2 answers:

V125BC [204]3 years ago

8 0

Answer:

you can see in above photo

Step-by-step explanation:

hope it helps!!

Neporo4naja [7]3 years ago

3 0

Answer:

(8p+9q)⋅(8p−9q)

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Can someone help me with this? 13 points!

zysi [14]

depth at point B = 5 m

Step-by-step explanation:

y = -(2/5)x - 1

Point C is located 10 m from the origin O which means that x = 10. Using this value in our equation,

y = -(2/5)(10) - 1

= -4 - 1

= -5

7 0

3 years ago

Four more than one third x

Ierofanga [76]

Answer:

4 + x/3

4 + 1/3x

Step-by-step explanation:

Step 1: Convert "four" to number

Step 2: Covert "more than" to number

4 +

Step 3: Convert "one-third" to number

4 + 1/3

Step 4: Add variable <em>x</em> (multiplication)

4 + 1/3x

5 0

3 years ago

Find the following Euler Totients using Euler’s Theorem, as explained on p.409 of the text (10 points each): a.ϕ(13) b.ϕ(81) c.ϕ

ad-work [718]

(a) $\varphi(13)=12$ since 13 is prime.

(b) $81=3^4$ , and there are 81/3 = 27 multiples of 3 between 1 and 81, which leaves 81 - 27 = 54 numbers between 1 and 81 that are coprime to 81, so $\varphi(81)=54$ .

(c) $100=2^2\cdot5^2$ ; there are 50 multiples of 2, and 20 multiples of 5, between 1 and 100; 10 of these are counted twice (the multiples of 2*5=10), so a total of 50 + 20 - 10 = 60 distinct numbers not coprime to 100, leaving us with $\varphi(100)=100-60=40$ .

(d) $102=2\cdot3\cdot17$ ; there are 51 multiples of 2, 34 multiples of 3, and 6 multiples of 17, between 1 and 102. Among these, we double-count 17 multiples of 2*3=6, 3 multiples of 2*17=34, and 2 multiples of 3*17=51; we also triple-count 1 number, 2*3*17=102. There are then 51 + 34 + 6 - (17 + 3 + 2) + 1 = 70 numbers between 1 and 102 that are not coprime to 102, and so $\varphi(102)=102-70=32$ .