The first way:
1. Reflect the figure I across the x-axis.
2. Translate the figure I and its reflection and rotate them 90 degrees.
The second way:
1. Reflect the figure I across the x-axis.
2. Rotate 90 degrees counterclockwise.
3. Rotate it across the y-axis.
0Answer:
A
Step-by-step explanation:
Find the zeros by letting y = 0 , that is
x² - x - 6 = 0 ← in standard form
(x - 3)(x + 2) = 0 ← in factored form
Equate each factor to zero and solve for x
x - 3 = 0 ⇒ x = 3
x + 2 = 0 ⇒ x = - 2
Since the coefficient of the x² term (a) > 0
Then the graph opens upwards and will be positive to the left of x = - 2 and to the right of x = 3 , that is in the intervals
(-∞, - 2) and (3, ∞ ) → option A
In this equation, p=27 because it is the last term and q=3 because it is the coefficient of the first term. First find all divisors of these numbers:
27: 1, 3, 9, 27
3:1, 3
Then use +/- p/q to find all possible combinations
+/-1, +/-3, +/-9, +/-27, +/- 1/3
Or your answer can be written as: 1, -1, 3, -3, 9, -9, 27,-27, 1/3, -1/3
Hope this helps!
X = 1/4, 5, -8
Hope this helps!
Answer:
5
Step-by-step explanation:
The 32 that have blue and green ribbons include the 16 that have all three, so there are only 32 -16 = 16 that have only blue and green ribbons.
The 31 that have green and white ribbons likewise include the 16 with all three, so there are only 31 -16 = 15 that have only green and white ribbons.
The 38 that have blue and white ribbons include the 16 with all three, so there are only 38 -16 = 22 that have only green and white ribbons.
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If we add the numbers of blue, green, and white ribbons, we are counting twice the numbers that have 2 ribbons, and 3 times the numbers that have 3 ribbons. We want to count each kind of ribbon-holder only once. Hence the number of individual dogs with any number of ribbons is only ...
62 +55 +63 -(16 +15 +22) -2(16) = 95
Of the 100 dogs, 95 have ribbons, so 5 dogs have not learned any tricks.
