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klasskru [66]
3 years ago
7

If f(x) is a continuous function defined for all real numbers, f(–10) = –2, f(–8) = 5, and f(x) = 0 for one and only one value o

f x, then which of the following could be that x value?
–7
–9
0
2

Mathematics
1 answer:
Vinvika [58]3 years ago
8 0
F(-10)=-2,  f(-8)=5, and f is continuous. All these mean that f(x) is 0 for x in between -10 and -8.

That is, f at -10 is negative, then at -8 is positive. So the graph of f "cuts" the x-axis for an x value in (-10, -8). 

Since there is only one value such that f(x)=0, x can only be -9.


Answer: -9

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<u>Given</u>:

The given figure consists of a triangle, a rectangle and a half circle.

The base of the triangle is 2 mi.

The height of the triangle is 4 mi.

The length of the rectangle is 9 mi.

The diameter of the half circle is 4 mi.

The radius of the half circle is 2 mi.

We need to determine the area of the enclosed figure.

<u>Area of the triangle:</u>

The area of the triangle can be determined using the formula,

A=\frac{1}{2}bh

where b is the base and h is the height

Substituting b = 2 and h = 4, we get;

A=\frac{1}{2}(2\times 4)

A=4 \ mi^2

Thus, the area of the triangle is 4 mi²

<u>Area of the rectangle:</u>

The area of the rectangle can be determined using the formula,

A=length \times width

Substituting length = 9 mi and width = 4 mi, we get;

A=9 \times 4

A=36 \ mi^2

Thus, the area of the rectangle is 36 mi²

<u>Area of the half circle:</u>

The area of the half circle can be determined using the formula,

A=\frac{\pi r^2}{2}

Substituting r = 2, we get;

A=\frac{(3.14)(2)^2}{2}

A=\frac{(3.14)(4)}{2}

A=\frac{12.56}{2}

A=6.28

Thus, the area of the half circle is 6.28 mi²

<u>Area of the enclosed figure:</u>

The area of the entire figure can be determined by adding the area of the triangle, area of rectangle and area of the half circle.

Thus, we have;

Area = Area of triangle + Area of rectangle + Area of half circle

Substituting the values, we get;

Area=4+36+6.28

Area = 46.28 \ mi^2

Thus, the area of the enclosed figure is 46.28 mi²

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