Given:
The graph of triangle PQR and triangle P'Q'R'.
To find:
The transformation that will map the triangle PQR onto P'Q'R'.
Solution:
From the given graph it is clear that the triangle PQR is formed in II quadrant and its base lies on the negative direction of x-axis.
The triangle P'Q'R' is formed in IV quadrant and its base lies on the positive direction of x-axis.
This is possible it the figure is rotated 180 degrees about the origin.
Therefore, the correct option is A.
For this case the main function is:
f (x) = x ^ 2
We are going to apply the following transformations:
Vertical translations
Suppose that k> 0:
To graph y = f (x) + k, move the graph of k units up.
We have then:
f (x) = x ^ 2 + 5
Horizontal translations
Suppose that h> 0
To graph y = f (x + h), move the graph of h units to the left.
We have then:
f (x) = (x + 1) ^ 2 + 5
Answer:
f (x) = (x + 1) ^ 2 + 5
Answer:
5/9 (F-32) = 25
Step-by-step explanation:
5/9 (F-32) = C
Multiply each side by 9/5
5/9 * 9/5 (F-32) =9/5 C
F -32 = 9/5 C
Add 32
F -32 +32 = 9/5 C +32
F = 9/5 C +32
We want 25 C to Fahrenheit
F = 9/5 *25 +32
F = 9*5 +32
F = 45+32
F = 77
(2,4), (4,8), (3,6), (5,10) because they’re all ratios of 1:2
Answer:
2x + 5y + 28 = 0
Step-by-step explanation:
since they are perpendicular,
m1 ×m2 = -1
5/2 × m2 = -1
m2 = -2/5
now,
y -y1 = M (x-x1)
y - (-4) = -2/5 ( x - (-4) )
y +4 = -2/5 ( x + 4 )
5 ( y +4 ) = -2 ( x+4)
5y +20 = -2x - 8
2x + 5y +20 + 8 =0
2x + 5y + 28 = 0
