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kicyunya [14]
3 years ago
10

Use the drip-down menus to complete the statements. A zero can be found between input values of blank because blank one zero of

the function is approximately blank

Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
8 0

Answer:

  • A zero can be found between input values of <u>   x = - 1 and x = 0     </u>  because one zero of the function is approximately <u>   - 0.5     </u>.

Explanation:

The first pictured attached contains the table with the data for this question.

The<em> input values</em> are the values of x.

Since the<em> function</em>, f(x), changes sign between x = - 1 (y = - 0.5) and x = 0 (y = 2.5), if you assume the function is continuous, you can assert that the graph of the function crosses the x-axis between those two values.

The x-coordinate where the graph crosses the x-axis is a zero of the function.

Then, you can assert that there is a zero between the input values of - 1 and 0.

That zero is approximately - 0.5. This is, the function crosses the x-axis approximately at x = - 0.5.

The second picture skecthes a graph showing the given points where you can see that the dotted curve crosses the x-axis close to x = - 0.5.

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Answer:

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Step-by-step explanation:

7.25% of 591.99 is 42.91, subtract 42.91 from 591.99 and the answer is 549.08, if the answer isn't there round the answer to the nearest possible answer.

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How many and what type of solution(s) does the equation have?
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2 irrational solutions.

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Find the two intersection points
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Answer:

Our two intersection points are:

\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)

Step-by-step explanation:

We want to find where the two graphs given by the equations:

\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1

Intersect.

When they intersect, their <em>x-</em> and <em>y-</em>values are equivalent. So, we can solve one equation for <em>y</em> and substitute it into the other and solve for <em>x</em>.

Since the linear equation is easier to solve, solve it for <em>y: </em>

<em />\displaystyle y = -\frac{3}{4} x + \frac{1}{4}<em />

<em />

Substitute this into the first equation:

\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16

Simplify:

\displaystyle (x+1)^2 + \left(-\frac{3}{4} x  + \frac{9}{4}\right)^2 = 16

Square. We can use the perfect square trinomial pattern:

\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16

Multiply both sides by 16:

(16x^2+32x+16)+(9x^2-54x+81) = 256

Combine like terms:

25x^2+-22x+97=256

Isolate the equation:

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We can use the quadratic formula:

\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

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\displaystyle x = \frac{-(-22)\pm\sqrt{(-22)^2-4(25)(-159)}}{2(25)}

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\displaystyle \begin{aligned} x &= \frac{22\pm\sqrt{16384}}{50} \\ \\ &= \frac{22\pm 128}{50}\\ \\ &=\frac{11\pm 64}{25}\end{aligned}

Hence, our two solutions are:

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We have our two <em>x-</em>coordinates.

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And:

\displaystyle y _2 = -\frac{3}{4}\left(-\frac{53}{25}\right) +\frac{1}{4} = \frac{46}{25}

Thus, our two intersection points are:

\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)

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