Answer:
frac{21x^6y^5}{14x^2y^9}
Factor the number =\frac{7\cdot \:3x^6y^5}{14x^2y^9}
Factor the number 14=7. 2 =\frac{7\cdot \:3x^6y^5}{7\cdot \:2x^2y^9}
Cancel\:the\:common\:factor:}\:7 =\frac{3x^6y^5}{2x^2y^9}
Step-by-step explanation:
Answer:
10=7
Step-by-step explanation:
6+4=10
2x5=10
10-3=7
Hope this helps :)
Answer: 31 : 9
Step-by-step explanation:
Assume the following:
Alice's amount = P
Bob's amount = Q
Amount received = n
If Alice receives $n$ dollars from Bob ;then she will have $4$ times as much money as Bob.
P + n = 4(Q - n)
P + n = 4Q - 4n
P = 4Q - 4n - n
P = 4Q - 5n - - - - (1)
If, on the other hand, she gives $n$ dollars to Bob, then she will have $3$ times as much money as Bob
P - n = 3(Q + n)
P - n = 3Q + 3n
P = 3Q + 3n + n
P = 3Q + 4n - - - - - - (2)
Equating both equations - (1) and (2)
4Q - 5n = 3Q + 4n
4Q - 3Q = 4n + 5n
Q = 9n
Express P in terms of n, use either equation (1) or (2)
From equation 2:
P = 3Q + 4n
Substituting Q = 9n
P = 3(9n) + 4n
P = 27n + 4n
P = 31n
Alice's amount = P, Bob's = Q
Ratio = P:Q
31 : 9
Question:
gbuhhubhhbuubbbubuubhbhhhbbubhbhbhubuhhbhbu
Answer:
yes
Exponentail thingies
easy, look at all them them, see that they have 5 in common?
rremember how esay it was to factor
ax^2+bx+c=0
now we have
5^(2x)-6(5^x)+5=0
remember that 5^(2x)=(5^2)^x or (5^x)^2
in other words, we can rewrite it as
1(5^x)^2-6(5^x)+5=0
if yo want, replace 5^x with a and factor
1a^2-6a+5=0
(a-1)(a-5)=0
a=5^x
(5^x-1)(5^x-5)=0
set each to zero
5^x-1=0
5^x=1
take the log₅ of both sides
x=log₅1
5^x-4=0
5^x=4
take the log₅ of both sides
x=log₅4
x=log₅1 and/or log₅4
second quesiton
same thing
1(2^x)-10(2^x)+16=0
factor
(2^x-8)(2^x-2)=0
set each to zero
2^x-8=9
2^x=8
x=3
2^x-2=0
2^x=2
x=1
x=3 or 1
first one
x=log₅1 and/or log₅4
second one
x=1 and/or 3
