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Bond [772]
1 year ago
11

Domain and range of linear and quadratic functions worksheet

Mathematics
1 answer:
Vlad [161]1 year ago
7 0

If the value of a is negative, then the range will be (-∞, k) and if the value of the a is positive then the range will be (k, ∞).

<h3>What is a quadratic equation?</h3>

It's a polynomial with a worth of nothing.

There exist polynomials of variable power 2, 1, and 0 terms.

A quadratic condition is a condition with one explanation where the degree of the equation is 2.

Domain and range of linear and quadratic functions

Let the linear equation be y = mx + c.

Then the domain and the range of the linear function are always real.

Let the quadratic equation will be in vertex form.

y = a(x - h)² + k

Then the domain of the quadratic function will be real.

If the value of a is negative, then the range will be (-∞, k) and if the value of the a is positive then the range will be (k, ∞).

More about the quadratic equation link is given below.

brainly.com/question/2263981

#SPJ4

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PLEASE HELP ME:<br><br> Find the perimeter of the figure. Round to the nearest tenth.
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Answer:

perimeter = 20.9 units

Step-by-step explanation:

perimeter

perimeter = distance around two dimensional shape

= addition of all sides lengths

<h2>perimeter of the figure</h2><h2>= AB+BC+CD+AD</h2>

distance formula:

d =  \sqrt{(  x_{2} -  x_{1})  {}^{2}   + ( y_{2} -  y_{1}) {}^{2}  }

<h3>1) distance of AB</h3>

A(-3,0) B(2,4)

x1 = -3 x2 = 2

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(substitute the values into the distance formula)

ab =  \sqrt{(2 - ( - 3)) {}^{2}  + (4 - 0) {}^{2} }

ab =  \sqrt{5 {}^{2} + 4 {}^{2}  }

ab =  \sqrt{41}AB = 6.4 units

<h3>2) distance of BC</h3>

B(2,4) C(3,1)

x1 = 2 x2 = 3

y1 = 4 y2 = 1

bc =  \sqrt{(3 - 2) {}^{2}  + (1 - 4) {}^{2} }

bc = \sqrt{1 {}^{2} + ( - 3) {}^{2}  }

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<h3>3) distance of CD</h3>

C(3,1) D(-4,-3)

x1 = 3 x2 = -4

y1 = 1 y2 = -3

cd =  \sqrt{( - 4 - 3) {}^{2}  + ( - 3 - 1)) {}^{2} }

cd =  \sqrt{( -   7) {}^{2}  + ( - 4 ){}^{2} }

cd =   \sqrt{65}

CD = 8.1 units

<h3>4) distance of AD</h3>

A(-3,0) D(-4,-3)

x1 = -3 x2 = -4

y1 = 0 y2 = -3

ad =  \sqrt{(  - 4 - ( - 3)) {}^{2}  +  ( - 3 - 0) {}^{2} }

ad =  \sqrt{( - 1) {}^{2} + ( - 3) {}^{2}  }

ad =  \sqrt{10}

AD = 3.2 units

<h2>perimeter of figure</h2>

= AB+BC+CD+AD

= 6.4 + 3.2 + 8.1 + 3.2

= 20.9 units

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