Alright, so we know that the race is 5 kilometers, so the equation will be 5-<some value>= <distance from finish line>. We also know that the student runs a kilometer every three minutes, so 3x=1km . Multiplying both sides by 5, we get 15y=5km (y being the number to make the equation make sense, or the slope). When the student has run 5km, the distance from the student to the finish line should be 0, so we get that 5-5=0, and plugging 15y in for 5 we get 5-15y=0. For 15x to equal 5, 3y=1 and y=1/3. Therefore, we plug that in for y, getting 5-15(1/3)=0. However, we have to make it for all times! Since 15 represents the minutes, we make that x, and since 0 represents the distance remaining, we make that the distance remaining, making it 5-(1/3)x=distance left. You can also think of y as the slope in y=mx+b - it stays constant that way.
Answer:
nth term = 1 1/2n -1
Step-by-step explanation:
The arithmetic sequence formula is:
a
n
=
a
1
+
(
n
–
1
)
d
Where:
a
n
is the nth term in the sequence
a
1
is the first term in the sequence
n
is the term you are solving for
d
is the common difference for any pair of consecutive numbers in the sequence.
First Term or
n
=
1
:
This is given in the problem.
a
1 = 9
Second Term or
n
=
2
:
Substitute
2 for n
in the formula and substitute the values from the problem giving:
a
2
=
9
+
(
(
2
–
1
)
×
-2
)
a
2
=
9
+
(
1
×
-2
)
a
2
=
9
+-2
a
2
=
7
Fifth Term or n
=
5
:
Substitute in the formula and substitute the values from the problem giving:
a
5
=
9+
(
(
5–
1
)
×
-2
)
a
5
=
9
+
(
4
×
-2
)
a
2
=
9
+
-8
a
2
= 1
Using this same process you should be able to determiner the
Third Term or n
=
3
: and Fourth Term or n
=
4
:
The height is 4 cm and the base is 8 cm.
Answer:
what do you need help with?
Step-by-step explanation:
Answer:
B) The heights of the bear should equal the class frequency.
Step-by-step explanation:
In drawing a histogram, the heights of the bear should equal the class frequency. as in histogram height of bar represent the frequency density and it´s area represent the frequency of class interval. Frequency distribution are represented by mean of rectangle. Earlier in the bar graph, width of bar does not represent any information, however, histogram´s bar width represent class interval.
