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nydimaria [60]
3 years ago
6

Expand and simplify :2(2x+3)-2(x-2) Pls help

Mathematics
2 answers:
Schach [20]3 years ago
8 0

Expand = 2⋅2⋅x+2⋅3−(2⋅x−2⋅2)

Simplify = 10+2⋅x

fgiga [73]3 years ago
8 0
The answer is 2x+10 hope it helps
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5/10=u/9

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A right triangle has one vertex on the graph of y = 9 - x^2 , x &gt; 0, at ( x , y ), another at the origin, and the third on th
jenyasd209 [6]

Answer:

  a.  A(x) = (1/2)x(9 -x^2)

  b.  x > 0 . . . or . . . 0 < x < 3 (see below)

  c.  A(2) = 5

  d.  x = √3; A(√3) = 3√3

Step-by-step explanation:

a. The area is computed in the usual way, as half the product of the base and height of the triangle. Here, the base is x, and the height is y, so the area is ...

  A(x) = (1/2)(x)(y)

  A(x) = (1/2)(x)(9-x^2)

__

b. The problem statement defines two of the triangle vertices only for x > 0. However, we note that for x > 3, the y-coordinate of one of the vertices is negative. Straightforward application of the area formula in Part A will result in negative areas for x > 3, so a reasonable domain might be (0, 3).

On the other hand, the geometrical concept of a line segment and of a triangle does not admit negative line lengths. Hence the area for a triangle with its vertex below the x-axis (green in the figure) will also be considered to be positive. In that event, the domain of A(x) = (1/2)(x)|9 -x^2| will be (0, ∞).

__

c. A(2) = (1/2)(2)(9 -2^2) = 5

The area is 5 when x=2.

__

d. On the interval (0, 3), the value of x that maximizes area is x=√3. If we consider the domain to be all positive real numbers, then there is no maximum area (blue dashed curve on the graph).

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3 years ago
Please helpppp! i’ll mark brainliest
Fiesta28 [93]

Answer:

from what i see i see thats its D

if not im sorry but i hope i got it right

Step-by-step explanation:

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