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3 years ago

Helpp, since this question is a bit hard, ill give brainliest and 20 pts.

Mathematics

1 answer:

Morgarella [4.7K]3 years ago

5 0

Step-by-step explanation:

A. When dealing with large numbers, sometimes it's easier to write them in a form of exponents of number ten. The value of the exponent shows how many times we multiply 10 by itself. That means that 10^3 is 10•10•10 or that 10^6 is 10•10•10•10•10•10.

So, when finding how many times on number is greater then the other, we need to divide them. We divide 8x10^6 by 2x10^5. It is done by dividing the numbers and subtracting the exponents; 8/2=4 and 10^6/10^5 is 10^6-5=10^1. So the correct answer is 4x10^1 which is 4x10, and that is 40.

B. Now, we have a total number of coins (2.25x10^5), and diametar of a coin (19mm = 19x10^-6km). Our task is to calculate the distance across which the coiks laid side-by-side would expand. We can find this by multiplying the number of coins with a diametar of single penny, 2.25x10^5•19x10^-6. Multiplying is done by multiplying the numbers (2.25x19=42.75) and adding the exponents (5+(-6)=5-6=-1). So, the distance is 42.75x10^-1km, which equals to 4.275 km. Obviously, this is less then the stated 5 km given in the text, so the reportet's statement is false.

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3 years ago

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3 years ago

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2 years ago

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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

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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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