5 3/5 divided by 2 2/3
1 answer:
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Answer: Yes, she will have enough trim for all four sides of the square, because the perimeter of the square photo (30.98 inches) is less than 32 inches of trim she has.
Step-by-step explanation:
The formula for calculate the area of a square is:
Where "s" is the lenght of any side of the square.
The formula for calculate the perimeter of a square is:
Where "s" is the lenght of any side of the square.
We know that that the area of the photo is 60 square inches, therefore, we can solve for "s" from the formula and find its value:
Substituting the value of "s" into the formula , we get that the perimeter of the photo is:
Therefore, since Alicia has 32 inches of trim and , we can conclude that she will have enough trim for all four sides of the square.
The equation is 
.
We are looking for a function with a vertex above the x-axis and a function that opens upward (has coefficient a > 0).
The first function opens downward and intersects the x-axis. The second function has a vertex below the x-axis. The third function satisfies our requirements. The fourth function has a vertex on the x-axis.
We can solve this algebraically with the knowledge that the real solutions of a quadratic are its x-intercepts. If there are no x-intercepts (because it lies entirely above or below the x-axis), then there are no real solutions. This is true when the discriminant
. You can see that from the quadratic formula. This holds true for both answers A and C, so to find the correct one, we remember that when the coefficient a of the 
term is positive, the graph opens upwards, so we choose C.
We are looking for a function with a vertex above the x-axis and a function that opens upward (has coefficient a > 0).
The first function opens downward and intersects the x-axis. The second function has a vertex below the x-axis. The third function satisfies our requirements. The fourth function has a vertex on the x-axis.
We can solve this algebraically with the knowledge that the real solutions of a quadratic are its x-intercepts. If there are no x-intercepts (because it lies entirely above or below the x-axis), then there are no real solutions. This is true when the discriminant