If a rectangle's length is y−2 and the width is 3y+2 write an expression for the perimeter and an expression for the area.
1 answer:
Answer:
Area = units square.
Perimeter = 8y units.
Step-by-step explanation:
Area of a rectangle = length * width.
Perimeter of a rectangle = 2*(length + width).
It is given that length = y−2 units and width = 3y+2 units. To find the area and the area of the rectangle in terms of y, simply put the length and the width in the above area and perimeter equations.
Area = length * width = (y-2)*(3y+2).
Expanding the expression gives:
Area = 3y^2 + 2y - 6y - 4 = (3y^2 -4y - 4) units square.
Similarly,
Perimeter = 2*(length + width) = 2*(y-2 + 3y+2) = 2*4y = 8y units.
Therefore, the expression for the area and the perimeter of the given rectangle is () units square and (8y) units respectively!!!
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Answer:
k = 11/3
Step-by-step explanation:
If the line is tangent to the curve, then Δ = 0.
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Remembering:
Δ>0 two different points of intersection x' x''
Δ=0 one point of intersection x' = x''
Δ<0 two different points of intersection in the complex plan x' and -x'
______________
As the line and the curve have one point of intersection, which is (x, y), we can make a equality between them:
2x + k = 3x² + 4
0 = 3x² - 2x + (4 - k)
Now we can use the Δ=0 (Δ= b² - 4ac)
Δ = 0 = (-2)² - 4.3.(4-k)
0 = 4 - 48 + 12k
12k = 44
k = 44/12 = 2 . 22 /3. 2.2 = 22/3.2 = 11/3
k = 11/3