Answer:
Step-by-step explanation:
log8 62 = x1
8^x1= 62
8^2 =64 >62 so <u>x1<2</u>
log7 50=x2
7^x2=50
7^2=49<50 so <u>x2>2</u>
we have, x1<2 and 2<x2
x1<2<x2
x1<x2
log8 62<log7 50
Answer: 1/70
Step-by-step explanation:
This is a question that can also be interpreted as what is the probability of having the first number of a phone number to be 8 and the last number of the phone number to also be 8. This answer gives the fraction of the phone numbers that starts with 8 and end with 8.
Since three numbers (0,1,2) cannot start a phone number and we are left to pick from 7 numbers,
then the probability of figure "8" starting phone number = 1/7
Since all 10 numbers can possibly end a phone number,
then the probability of having figure "8" as the last digit of a phone number = 1/10
Hence probability of having "8" as the first and last digit of a phone number = fraction of total telephone numbers that begin with digit 8 and end with digit 8 = 1/7 × 1/10 = 1/70.
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36 it the surface area of the triangler prism
Answer:
Step-by-step explanation:
In this particular case we have the following system of equations:
y
=
−
3
x
+
4
[
E
q
.
1
]
x
+
4
y
=
−
6
[
E
q
.
2
]
Substituting
[
E
q
.
1
]
in
[
E
q
.
2
]
:
x
+
4
(
−
3
x
+
4
)
=
−
6
Applying the distributive property on the left side:
x
−
12
x
+
16
=
−
6
Simplifying
:
−
11
x
=
−
22
Solving for
y
:
x
=
−
22
−
11
=
2
Substituting
x
=
2
in
[
E
q
.
1
]
:
y
=
−
3
(
2
)
+
4
=
−
2
Therefore
, the solutions are
x
=
2
and
y
=
−
2
