Answer:
c,d
Step-by-step explanation:
![ab^{-3x}=a(b^{-3})^x=a(\frac{1}{b^3} )^x=a[(\frac{1}{b} )^3]^x\\=a(\frac{1}{b} )^{3x}](https://tex.z-dn.net/?f=ab%5E%7B-3x%7D%3Da%28b%5E%7B-3%7D%29%5Ex%3Da%28%5Cfrac%7B1%7D%7Bb%5E3%7D%20%29%5Ex%3Da%5B%28%5Cfrac%7B1%7D%7Bb%7D%20%29%5E3%5D%5Ex%5C%5C%3Da%28%5Cfrac%7B1%7D%7Bb%7D%20%29%5E%7B3x%7D)
Answer:
option 2 is correct
Step-by-step explanation:
Answer:
(3x+4)(5x+7)
Step-by-step explanation:
15x^2
+41x+28
Factor the expression by grouping. First, the expression needs to be rewritten as 15x^2
+ax+bx+28. To find a and b, set up a system to be solved.
a+b=41
ab=15×28=420
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 420.
1,420
2,210
3,140
4,105
5,84
6,70
7,60
10,42
12,35
14,30
15,28
20,21
Calculate the sum for each pair.
1+420=421
2+210=212
3+140=143
4+105=109
5+84=89
6+70=76
7+60=67
10+42=52
12+35=47
14+30=44
15+28=43
20+21=41
The solution is the pair that gives sum 41.
a=20
b=21
Rewrite 15x^2
+41x+28 as (15x^2
+20x)+(21x+28).
(15x^2
+20x)+(21x+28)
Factor out 5x in the first and 7 in the second group.
5x(3x+4)+7(3x+4)
Factor out common term 3x+4 by using distributive property.
(3x+4)(5x+7)
Answer:
103&（#！92（$+#”39214849）？lm
D A A C A C A B A D A C B B 1/1/3 B 43/1/5 A A
