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strojnjashka [21]
3 years ago
9

if the zeros of a quadratic function , f, are -3 and 5, what is the x-coordinate of the vertex of the function f?

Mathematics
1 answer:
Blababa [14]3 years ago
8 0

we know function "f" has zeros at -3 and 5, namely x = -3 and x = 5, well for a quadratic, the vertex is half-way between those two zeros,

-3................... 1....................5

so the x-coordinate is at 1.  Check the picture below, regardless of the shape of the parabola.

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Veronika [31]

Answer:

x =147

Step-by-step explanation:

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42+(-124+x) =360-(2x+1)

Combine like terms

x-82 =359-2x

Add 2x to each side

x-82+2x = 359

3x -82 = 359

Add 82 to each side

3x-82+82 = 359+82

3x = 441

Divide by 3

3x/3=441/3

x =441/3  or 147

4 0
3 years ago
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3 years ago
Find the roots of h(t) = (139kt)^2 − 69t + 80
Sonbull [250]

Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

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Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

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The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

7 0
2 years ago
Evaluate the expression. 0.1×( – 0.9+ – 0.2÷ – 0.5) Write your answer as an integer or a decimal. Do not round.
matrenka [14]

Answer:

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Step-by-step explanation:

Given the expression :

0.1×( – 0.9+ – 0.2÷ – 0.5)

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Solving the division first

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Now we have ;

0.1 * (-0.9 + 0.4)

0.1 * - 0.5

= - 0.05

4 0
2 years ago
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