Actually, this answer would be true. Why?
The first equation is: a(sub <em>n</em>) = 8, 13, 18, 23
The second is: a(sub 1)=8 ; a(sub <em>n</em>)= a(sub <em>n</em>-1)+5
if you wish to find the second term, plug two into the equation for <em /><em>n</em>
8+5=13
to find the third, plug the second term, 13, in for <em>n.</em>
13+5=18.
Hope this helped! I know it's a bit on the late side, but at least you can get the general idea!
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
Answer:
3rd choice
Step-by-step explanation:
the 3rd choice is the same as all of the others just in fraction form .you need to keep fraction form for the statement to still be true
We have learned that, in in an algebraic expression, letters can stand for numbers. When we substitute a specific value for each variable, and then perform the operations, it's called evaluating the expression. Let's evaluate the expression 3y + 2y when 5 = y.
X • 1.5 = 300
So...
300/ 1.5 = 200
He planted 200 acres last year