The width of the rug is 3 inches
Step-by-step explanation:
A knitter wants to make a rug for a dollhouse
- The length of the rug will be 2 inches more than it’s width
- The total area of the rug ( in square inches) based on the width w of the rug (in inches ) is given by A(w) = w (2 + w)
- The desired area of the rug is 15 square inches
We need to find the width of the rug
∵ The area of the rug is A(w) = w(2 + w)
- Simplify the right hand side
∴ A(w) = 2w + w²
∵ The desired area of the rug is 15 square inches
- Equate A(w) by 15
∴ 2w + w² = 15
- Subtract 15 from both sides
∴ 2w + w² - 15 = 0
- Its a quadratic equation arrange its terms from the greatest power of x
∴ w² + 2w - 15 = 0
Factorize it into two binomial factors
∵ w² = w × w
∵ 15 = 3 × 5
∵ 5w - 3w = 2w
∴ w² + 2w - 15 = (w -3)(w + 5)
∴ (w - 3)(w + 5) = 0
- Equate each factor by 0
∴ w - 3 = 0 OR w + 5 = 0
∵ w - 3 = 0
- Add 3 to both sides
∴ w = 3
∵ w + 5 = 0
- Subtract 5 from both sides
∴ w = -5 ⇒ rejected because no dimension with -ve value
∴ w = 3 only
The width of the rug is 3 inches
Learn more:
You can learn more about the quadratic equation in brainly.com/question/9328925
#LearnwithBrainly
from the question, (-2,2) and (1,2) have the same y value so you can use that as your base and easily find the perpendicular height using the y axis since it's parallel to the x axis.
the third point, (0,-6), to the base is your height
use the sum of the positive of the y values to find your height (because height can't be negative): 6+2 = 8
area of a triangle = 1/2 bh = 1/2 x 3 x 8 = 12
Note: 1/2 bh only works because (-2,2) and (1,2) form a line parallel to the x axis
Answer:
y-3=0 (x+6)
Step-by-step explanation:
Hope im right :)
The equation of a circle exists:
, where (h, k) be the center.
The center of the circle exists at (16, 30).
<h3>What is the equation of a circle?</h3>
Let, the equation of a circle exists:
, where (h, k) be the center.
We rewrite the equation and set them equal :
We solve for each coefficient meaning if the term on the LHS contains an x then its coefficient exists exactly as the one on the RHS containing the x or y.
-2hx = -32x
h = -32/-2
⇒ h = 16.
-2ky = -60y
k = -60/-2
⇒ k = 30.
The center of the circle exists at (16, 30).
To learn more about center of the circle refer to:
brainly.com/question/10633821
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