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Sati [7]
4 years ago
8

Write a quadratic function h whose zeros are 5 and -7

Mathematics
1 answer:
marin [14]4 years ago
3 0

Quadratic function whose zeros are 5 and -7 is x^2+2x-35=0

Step-by-step explanation:

Write a quadratic function h whose zeros are 5 and -7

If 5 and -7 are zeros of quadratic function then:

x=5 and x=-7

or x-5 = 0 and x+7=0

So, Finding quadratic equation:

(x-5)(x+7)=0\\x(x+7)-5(x+7)=0\\x^2+7x-5x-35=0\\x^2+2x-35=0

So, quadratic function whose zeros are 5 and -7 is x^2+2x-35=0

Keywords: Quadratic Function

Learn more about Quadratic Function at:

  • brainly.com/question/7361044
  • brainly.com/question/9328925
  • brainly.com/question/1357167

#learnwithBrainly

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3

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2/3+2/3+2/3=2

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NEED HELP ON THIS ASAP!
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Answer: 40/9

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Step-by-step explanation:

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Find, correct to the nearest degree, the three angles of the triangle with the vertices d(0,1,1), e( 2, 4,3) − , and f(1, 2, 1)
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Well, here's one way to do it at least... 

<span>For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB). </span>

<span>Let P=(4,0) be the projection of B onto the x-axis. </span>
<span>Let Q=(-3,0) be the projection of C onto the x-axis. </span>

<span>Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan(5/4). </span>

<span>Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan(2). </span>

<span>Angle A, then, is 180 - atan(5/4) - atan(2) = 65.225. One down, two to go. </span>

<span>||b|| = sqrt(41) (use Pythagorian Theorum on triangle AQC) </span>
<span>||c|| = sqrt(45) (use Pythagorian Theorum on triangle APB) </span>

<span>Using the Law of Cosines... </span>
<span>||a||^2 = ||b||^2 + ||c||^2 - 2(||b||)(||c||)cos(A) </span>
<span>||a||^2 = 41 + 45 - 2(sqrt(41))(sqrt(45))(.4191) </span>
<span>||a||^2 = 86 - 36 </span>
<span>||a||^2 = 50 </span>
<span>||a|| = sqrt(50) </span>

<span>Now apply the Law of Sines to find the other two angles. </span>

<span>||b|| / sin(B) = ||a|| / sin(A) </span>
<span>sqrt(41) / sin(B) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(41) / sqrt(50) = sin(B) </span>
<span>.8222 = sin(B) </span>
<span>asin(.8222) = B </span>
<span>55.305 = B </span>

<span>Two down, one to go... </span>

<span>||c|| / sin(C) = ||a|| / sin(A) </span>
<span>sqrt(45) / sin(C) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(45) / sqrt(50) = sin(C) </span>
<span>.8614 = sin(C) </span>
<span>asin(.8614) = C </span>
<span>59.470 = C </span>

<span>So your three angles are: </span>

<span>A = 65.225 </span>
<span>B = 55.305 </span>
<span>C = 59.470 </span>
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