Answer:
2(n+1)+2
You start with two greens and two columns of two orange squares while adding two orange squares each time. So, the bolded part is the green squares that stay the same. The 2(n+1) represents the two orange columns that increase by one block on each side per image.
Answer:
expected profit is - $0.50
Step-by-step explanation:
1 $(7.00) 0.166666667 $(1.17)
2 $(7.00) 0.166666667 $(1.17)
3 $1.00 0.166666667 $0.17
4 $1.00 0.166666667 $0.17
5 $1.00 0.166666667 $0.17
6 $8.00 0.166666667 $1.33
$(0.50)
The true statement about the set of quadrilaterals in the coordinate plane is B. Because quadrilateral ABCD can be reflected across the x -axis, then rotated 90° counterclockwise about the origin, and then dilated about the origin by a scale factor of 2 to obtain quadrilateral A'B'C'D' , then quadrilateral ABCD is similar to quadrilateral A'B'C'D'.
<h3>What is a quadrilateral?</h3>
A quadrilateral is a polygon having four sides, four angles, and four vertices.
In this case, because quadrilateral ABCD can be reflected across the x -axis, then rotated 90° counterclockwise about the origin, and then dilated about the origin by a scale factor of 2 to obtain quadrilateral A'B'C'D' , then quadrilateral ABCD is similar to quadrilateral A'B'C'D'.
Learn more about quadrilateral on:
brainly.com/question/23935806
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Answer:
Step-by-step explanation:
Solving for x means you have to factor. First factor out the GCF of 2 to get:
and now we'll factor using the regular old method of ac and then factoring by grouping. In our polynomial, a = 3, b = 1, c = -6. Therefore, a times c is 3 * -6 which is -18. We need some combinations of the factors of 18 that will add to give us 1, the b term in the middle. The factors of 18 are:
1, 18
2, 9
3, 6 and that's it. Hm...it seems that won't work, so let's throw this into the quadratic formula, going back to the original and a = 6, b = 2 and c = -12:
and
and
and
and
which finally simplifies to
No wonder that didn't factor using the traditional method of factoring! We could have found that out by finding first the value of the discriminant, but oh well!