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Darina [25.2K]
3 years ago
10

The table of values represents the function g(x) and the graph shows the function f(x).

Mathematics
1 answer:
snow_tiger [21]3 years ago
8 0
<h2>Hello!</h2>

The answer is:

The first and second options:

f(x) and g(x) intersect at exactly two points.

The x-intercepts of f(x) are common to g(x)

<h2>Why?</h2>

To find the correct option (or options) , we need to remember the following:

- When a function intercepts the y-axis, it means that the "x" coordinate will be equal to 0.

- When a function intercepts the x-axis, it means that the "y" coordinate will be equal to 0.

Now, to find the correct option, we also need to compare the graphed function (f(x))  to the given table (g(x)).

So, discarding each of the given options to find the correct option, we have:

- First option, f(x) and g(x)  intersect at exactly two points: True.

From the graph we can see that f(x) intercepts the x-axis at two points (-1,0) and (1,0), also, from the table we can see that g(x) intercepts the x-axis at the same two points (-1,0) and (1,0), it means that the functions intersect at exactly two points.

Hence,  we have that f(x) and g(x)  intersect at exactly two points.

- Second option, the x-intercepts of f(x) are common to g(x): True.

From the graph we can see that f(x) intercepts the x-axis at two points (-1,0) and (1,0), also, from the table we can see that g(x) intercepts the x-axis at the same two points (-1,0) and (1,0), so, both functions intercepts the x-axis at common points.

Hence,we have that the x-intercepts of f(x) are common to g(x)

- Third option, he minimum value of f(x) is less than the minimum value of g(x): False.

From the graph, we can see that the minimum value of f(x) is located at the point (0,-1), also, from the given table for g(x) we can see that there are values below the point (2,-3), meaning that the minimum value of f(x) is NOT less than the minimum value of g(x).

Hence, we have that the minimum value of f(x) is NOT less than the minimum value of g(x).

- Fourth option, f(x) and g(x) have the same y-intercept: False.

We can see that for the function f(x) the y-intercept is located at (0,-1) while from the given table, we can see the y-intercept for the function g(x) is located at (0,1)

Hence,  we have that f(x) and g(x) have differents y-intercepts.

Therefore, the correct answers are:

The first and second options:

f(x) and g(x) intersect at exactly two points.

The x-intercepts of f(x) are common to g(x)

Have a nice day!

Note: I have attached an image for better understanding.

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The right-side ratio of (5+9)/9 = 14/9 suggests that BC 9 is 14/9 times the length DE. If that were the case, BC = (14/9)(11) = 154/9 = 17 1/9 ≈ 17.1, as Toad says.

The different ratios of the two sides (3/2 vs 14/9) tell you that the triangles are NOT similar, so the length of BC cannot be found by referring to the ratios of the given sides.

Rather, the Law of Cosines must be invoked, first to find angle A (109.471°), then to use that angle to compute the length of BC given the side lengths AB and AC. That computation gives BC ≈ 16.971. (See the second attachment.)

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A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,
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Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of \pi/4.

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

d(\theta)=\frac{v_i^2sin(2\theta)}{g}

Where v_i is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is [0, \pi/2]. The critical  points of the function are those who make d'(\theta)=0:

d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}

d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...

The critical value inside the interval is \pi/4.

d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft

The second step is to find the values of the function at the endpoints of the interval:

d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft

The biggest value of f is gived by \pi/4, therefore \pi/4 is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of \pi/4.

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