NEED HELP IN SOLVING THIS AREA PROBLEM !!!!!

Mathematics

2 answers:

dezoksy [38]3 years ago

6 0

I’m not sure if it’s add or multiple the sides of each I don’t know if this even helped lol

disa [49]3 years ago

6 0

Answer:

796

Step-by-step explanation:

First turn it into 3 Composite Shapes, Then add the sums up to get you answer. You would separate them using the given numbers. DO NOT USE THE LARGE NUMBER: 53 you have to put them into 3 tiny squares using the numbers it gives you. Look at the image, to look at how to separate them and then i will continue the math.

To Do This You Also Have To Follow PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right)

I color coded it for you so that you could tell what block was what.

I also put color under the numbers that were also with the block that you needed to multiply.

For the first square/red square you need to do the math:

25x17=415

For the second square/blue square you need to do the math:

24x(25-12)=

24x13=312

For the third and final square/green square you need to do the math:

6x12=72

For the last step you have to add all the sums together to get the quotient/answer:

415+312+72=796

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A right triangle has one vertex on the graph of y = 9 - x^2 , x > 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).