Answer:
14, 28, 42, 56, and 70 are the first 5 common multiples of 14
20, 40, 60, 80, and 100 are the first 4 common multiples of 20
Answer:
oh no i forgor about this i just learned this in class :(
a.)
784
/ \
2 392
/ \
2 196
/ \
2 98
/ \
2 49
/ \
7 7
√784 = 7² × 2^4
= 7² × 4²
= <u>28²</u>
So, √784 = 28
b.)
2025
/ \
3 675
/ \
3 225
/ \
3 75
/ \
3 25
/ \
5 5
2025 = 5² × 3^4
= 5² × 9²
= <u>4</u><u>5</u><u>²</u>
So, √2025 = 45
c.)
9261
/ \
3 3087
/ \
3 1029
/ \
3 343
/ \
7 49
/ \
7 7
9261 = 7³ × 3³
= <u>21³</u>
So, ³√9261 = 21
d.)
46656
/ \
2 23328
/ \
2 11664
/ \
2 5832
/ \
2 2916
/ \
2 1458
/ \
2 729
/ \
3 243
/ \
3 81
/ \
3 27
/ \
3 9
/ \
3 3
46.656 = 2³ × 3³ × 2³ × 3³
= 6³ × 6³
= <u>3</u><u>6</u><u>³</u>
So, ³√46.656 = 36
<em>Hope it helps and is useful</em><em> </em><em>:</em><em>)</em>
Your answer would be 4x-6y-12
Answer:
p = ±21
Step-by-step explanation:
Given z1 = p +2i and z2 = 1 – 2i, If |z1/z2| = 13;
|p+2i/1-21| = 13
To get p from the expression above, we need to rationalize the complex function first.
p+2i/1-21 = p+2i/1-2i * 1+2i/1+2i
= p+2pi+2i+4i²/1-4i²
Since i² = -1;
= p+2pi+2i-4/1+4
= p-4+2i(p+1)/5
= p-4/5 + 2(p+1)/5 i
Then we will take the modulus of the resulting expression and equate to the value of 13 to get p
|p+2i/1-21| = √(p-4/5)²+ (2p+2/5)² = 13
(p-4/5)²+ (2p+2/5)² = 13²
(p-4)²+(2p+2)² = 13²*5²
p²-8p+16+4p²+8p+4 = 4225
5p²+20 = 4225
5p² = 4205
p² = 841
p = ±√841
p = ±21
The possible values of p are 21 and -21