Right triangle [tex]A B C[/tex] is located at [tex]A(-1,-2), B(-1,1)[/tex], and [tex]C(3,1)[/tex] on a coordinate plane. What is the equation of a circle [tex]A[/tex] with radius [tex]\overline{A C}[/tex]?
[tex](x+1)^2+(y+2)^2=9[/tex]
[tex](x+1)^2+(y+2)^2=25[/tex]
[tex](x-3)^2+(y-1)^2=16[/tex]
[tex](x-3)^2+(y-1)^2=25[/tex]
Answer (3)
Identify the width as x and represent the length of a rectangle as 2x + 5. With the area formula, set up the equation x(2x + 5) = 403. Solve this quadratic equation to find the value of x, which is the width of the rectangle. ;
The width of the rectangle is approximately 12.97 centimeters.
Let's denote:
W as the width of the rectangle (in centimeters),
I as the length of the rectangle (in centimeters).
According to the given information:
The length of the rectangle is 5 centimeters longer than twice the width: l = 2w + 5.
The area of the rectangle is 403 square centimeters: A = lw = 403
We have two equations here. We can use these equations to find the value of W, the width of the rectangle.
Substitute the expression for I from the first equation into the second equation:
w(2w + 5) = 403
Now, let's expand and rearrange the equation:
2 w 2 + 5 w − 403 = 0
Now, we have a quadratic equation. We can solve it using the quadratic formula:
w = 2 a − b ± b 2 − 4 a c = where a = 2 , b = 5 , and c = − 403 .[/tex]
Plugging in the values:
[tex] w = 2 × 2 − 5 ± 5 2 − 4 × 2 × ( − 403 ) w = 4 − 5 ± 25 + 3212 w = 4 − 5 ± 3237 w = 4 − 5 ± 56.89
We discard the negative root since width can't be negative.
So,
w = 4 − 5 + 56.89 w = 4 51.89 w ≈ 12.97
Rounding to two decimal places, the width of the rectangle is approximately 12.97 centimeters.
The width of the rectangle is approximately 13 centimeters, derived from the relationship between the width and length based on the area given. By setting up a quadratic equation and solving it, we determined that the valid solution for the width is 13 cm. The negative solution was discarded as width cannot be negative.
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