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Vedmedyk [2.9K]
1 year ago
6

Show that the function has at least one zero between x=1 and x=2

Mathematics
1 answer:
ANTONII [103]1 year ago
7 0

Answer:

See Below.

Step-by-step explanation:

We are given the function:

\displaystyle f(x) = 7x^5-9x^4-x^2

And we want to show that it has at least one zero between <em>x</em> = 1 and <em>x</em> = 2.

Because the function is a polynomial, it is everywhere continuous.

Evaluate the function at <em>x</em> = 1 and <em>x</em> = 2:

\displaystyle \begin{aligned} f(1) & = 7(1)^5 - 9(1)^4 - (1)^2 \\ \\ & = -3\end{aligned}

And:

\displaystyle \begin{aligned} f(2) & = 7(2)^5 - 9(2)^4 - (2)^2 \\ \\ & = 76 \end{aligned}

Therefore, because the function changes signs from <em>x</em> = 1 to <em>x</em> = 2 and is continuous on the interval [1, 2], by the intermediate value theorem, there must exist at least one zero in the interval.

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Which equation shows the quadratic formula used correctly to solve 5x2 + 3x – 4 = 0 for x? x = StartFraction negative 3 plus-or-
Brrunno [24]

Answer: FIRST OPTION

Step-by-step explanation:

<h3> The missing picture is attached.</h3>

By definition, given a Quadratic equation in the form:

ax^2+bx+c=0

Where "a", "b" and "c" are numerical coefficients and "x" is the unknown variable, you caN use the Quadratic Formula to solve it.

The Quadratic Formula is the following:

x=\frac{-b \±\sqrt{b^2-4ac} }{2a}

In this case, the exercise gives you this Quadratic equation:

5x^2 + 3x - 4 = 0

You can identify that the numerical coefficients are:

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You can identify that the equation that shows the Quadratic formula used correctly to solve the Quadratic equation given in the exercise for "x", is the one shown in the First option.

6 0
3 years ago
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4.) Which equations below have no solution? Select all that apply, and justify your response with work. a.) x - 9 = 2(x - 3) + 1
Sophie [7]

Answer:  Equations (b) and (e) do not have solutions

Step-by-step explanation:  b)  The x cancels out (-10x + 35) = -10x + 318;  35 = 318???? NO.  e)  3(x + 2) + 1 = x + 2(4 + x)​;  3x + 7 = 3x + 8.  The x cancels out again.

The other s can be solved for x.

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Find the x-intercept of the parabola of with vertex (1,20) and the y-intercept (0,16). write your answer in this form: (x1,y1),(
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I assume that the parabola in this particular problem is one whose axis of symmetry is parallel to the y axis. The formula we're going to use in this case is (x-h)2=4p(y-k). We know variables h and k from the vertex (1,20) but p is not given. However, we can solve for p by substituting values x and y in the formula with the y-intercept:

(0-1)^2=4p(16-20)

Solving for p, p=-1/16.

Going back to the formula, we can finally solve for the x-intercepts. Simply fill in variables p, h and k then set y to zero:

(x-1)^2=4(-1/16)(0-20)
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Here, we have two values of x

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thus, the answers are: (sqrt(5)+1,0) and (-sqrt(5)+1,0).
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elena-s [515]

the unit rate is 4:6

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The school student council sponsor a switch day where students were able to switch classes every 20 minutes the students are in
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60*7=420/20=21 the anwser is 21 
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