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Given that th<span>e coordinates of the vertices of △DEF are D(2, −1) , E(7, −1) , and F(2, −3) and the coordinates of the vertices of △D′E′F′ are D′(0, −1) , E′(−5, −1) , and F′(0, −3) .
Notice that the y-coordinates of the pre-image and that of the image are the same, which means that there is a reflection across the y-axis.
A refrection across the y-axis results in the change in sign of the x-coordinates of the pre-image and the image while the y-coordinate of the image remains the same as that of the pre-image.
A refrection across the y-axis of </span>△DEF with vertices D(2, −1) , E(7, −1) , and F(2, −3)
will result in and image with vertices (-2, -1), (-7, -1) and (-2, -3) respectively.
Notice that the x-coordinate of the final image △D′E′F′ with vertices <span>D′(0, −1) , E′(−5, −1) , and F′(0, −3) is 2 units greater than the vertices of the result of recting the pre-image across the y-axis.
This means that the result of refrecting the pre-image was shifted two places to the right.
Therefore, </span>the sequence of transformations that maps △DEF to △D′E′F′ are reflection across the y-axis and translation 2 units right.
Answer:
width =19 inches
length= 27 inches
Step-by-step explanation:
Let width be w inches so according to given scenario as length is 8 more than width so the equation becomes
l=8+w inches (Equation 1)
Area of Rectangle= L*w
Area= 513
L*w=513 (Equation 2)
Putting l's value from equation 1 in equation 2
(8+w)w=513
w^2+8w-513=0
Making Factors
w^2+27w-19w-513=0
w(w+27)-19(w+27)=0
(w-19)(w+27)=0
w=19, w=-27
as width cannot be negative so if we only consider w=19 then
l=8+19
l=27
The first one is the second one. t(n)=20n+480
Question 4: is the first one t(n)=8n+15
