Answer:
see explanation
Step-by-step explanation:
A difference of two squares is factored as
a² - b² = (a - b)(a + b)
(g)
1 - 81y²
= 1² - (9y)²
= (1 - 9y)(1 + 9y)
(j)
64x² - 25y²
= (8x)² - (5y)²
= (8x - 5y)(8x + 5y)
Answer:
graph{3x+5 [-10, 10, -5, 5]}
x
intercept:
x
=
−
5
3
y
intercept:
y
=
5
Explanation:
For a linear graph, the quickest way to sketch the function is to determine the
x
and
y
intercepts and draw a line between the two: this line is our graph.
Let's calculate the
y
intercept first:
With any function,
y
intercepts where
x
=
0
.
Therefore, substituting
x
=
0
into the equation, we get:
y
=
3
⋅
0
+
5
y
=
5
Therefore, the
y
intercept cuts through the point (0,5)
Let's calculate the
x
intercept next:
Recall that with any function:
y
intercepts where
x
=
0
.
The opposite is also true: with any function
x
intercepts where
y
=
0
.
If we substitute
y
=
0
, we get:
0
=
3
x
+
5
Let's now rearrange and solve for
x
to calculate the
x
intercept.
−
5
=
3
x
−
5
3
=
x
Therefore, the
x
intercept cuts through the point
(
−
5
3
,
0
)
.
Now we have both the
x
and
y
intercepts, all we have to do is essentially plot both intercepts on a set of axis and draw a line between them
The graph of the function
y
=
3
x
+
5
:
graph{3x+5 [-10, 10, -5, 5]}
2x + 3y/4 = 5....multiply everything by 4
8x + 3y = 20
3y = -8x + 20
y = -8/3x + 20/3 or y = (-8x + 20) / 3
Answer:
1. D. 20, 30, and 50
2. A. 86
3. B. 94
Step-by-step explanation:
1. To find the outliers of the data set, we need to determine the Q1, Q3, and IQR.
The Q1 is the middle data in the lower part (first 10 data values) of the data set (while the Q3 is the middle data of the upper part (the last 10 data values) the data set.
Since it is an even data set, therefore, we would look for the average of the 2 middle values in each half of the data set.
Thus:
Q1 = (85 + 87)/2 = 86
Q3 = (93 + 95)/2 = 94
IQR = Q3 - Q1 = 94 - 86
IQR = 8
Outliers in the data set are data values below the lower limit or above the upper limit.
Let's find the lower and upper limit.
Lower limit = Q1 - 1.5(IQR) = 86 - 1.5(8) = 74
The data values below the lower limit (74) are 20, 30, and 50
Let's see if we have any data value above the upper limit.
Upper limit = Q3 + 1.5(IQR) = 94 + 1.5(8) = 106
No data value is above 106.
Therefore, the only outliers of the data set are:
D. 20, 30, and 50
2. See explanation on how to we found the Q1 of the given data set as explained earlier in question 1 above.
Thus:
Q1 = (85 + 87)/2 = 86
3. Q3 = (93 + 95)/2 = 94
Answer:
you actually don't need a graph to be able to solve this equation. As long as you are given two data points, you can already solve for the linear equation. The most common way of writing linear equations is the slope-intercept form: y = mx + b, where m is the slope and b is the y-intercept.
Step-by-step explanation:
