Given J(1, 1), K(3, 1), L(3, -4), and M(1, -4) and that J'(-1, 5), K'(1, 5), L'(1, 0), and M'(-1, 0). What is the rule that tran
anastassius [24]
(x; y) -> (x - 2; y + 4)
J(1; 1) ⇒ J'(1 - 2; 1 + 4) = (-1; 5)
K(3; 1) ⇒ K'(3 - 2; 1 + 4) = (1; 5)
L(3;-4) ⇒ L'(3 - 2; -4 + 4) = (1; 0)
M(1;-4) ⇒ M'(1 - 2;-4 + 4) = (-1; 0)
Not likely because in my class we have more boys then girls
m=27
f=5
Answer:
Option C
Step-by-step explanation:
When you put the write the scenario in an equation, it'll look like
(5q+3)(5q-3) for the new width times the new length
Factor these, and you shall get 25q^2+15q-15q-9. Simplify that into
25q^2-9
Answer: The solution is the ordered pair (1/2, -3/4) so x = 1/2 and y = -3/4 pair up together. The two lines, when graphed, cross at this location.
note: 1/2 = 0.5 and -3/4 = -0.75; so the intersection point can be written as (0.5, -0.75)
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Explanation:
The second equation has y isolated. We can replace the y in the first equation with the expression 1/2x - 1
3x + 2y = 0
3x + 2( y ) = 0
3x + 2( 1/2x - 1) = 0 ... y is replaced with 1/2x-1
3x + 2(1/2x) + 2(-1) = 0 .... distribute
3x + x - 2 = 0 .... note how 2 times 1/2 is 1
4x - 2 = 0
4x = 2
x = 2/4
x = 1/2
Use this x value to find the y value it pairs with
y = (1/2)*x - 1
y = (1/2)*(1/2) - 1 ... replace x with 1/2
y = 1/4 - 1
y = 1/4 - 4/4
y = (1-4)/4
y = -3/4
Step-by-step explanation:
y iterscept is -4 slope 2/3 rise over run
