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2 years ago

Write a quadratic function f whose zeros are -10 and -2

Mathematics

2 answers:

statuscvo [17]2 years ago

7 0

$f(x) = (x - -10)(x - -2)$

We could stop here and have answered the question.

$f(x) = (x+10)(x+2) = x^2 + 12 x + 20$

Answer: $f(x)=x^2+12x + 20$

PtichkaEL [24]2 years ago

3 0

Hiiiiiiiiiiiiiiii mate......

here is your answer......✌✌✌✌

let alpha = -10 and beeta = -2
sum of zeroes= (alpha+beeta)=-10+-2=-12
product of zeroes= alpha.beeta= -10×-2=20
so,the required polynomial is

x^2-(alpha+beeta)X+alpha.beeta
x^2+12x+20

hope it help you...

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The length of a rectangle is 4 ya more than twice the width x. The area is 510 yd^2. Find the dimensions of the given shape

Furkat [3]

Answer:

The dimensions of the rectangle are

length = 34 yards

width = 15 yards

Step-by-step explanation:

Given:

The shape is of rectangle,

Width of rectangle = x

Area of rectangle = 510 yd²

According to the given condition

Length of rectangle = 4 + 2x

To Find:

length =?

Width = ?

Solution:

We know that,

$\textrm{Area of rectangle} = Length\times Width$

On substituting the given values we get

$510 = (4 + 2x)\times x\\510 = 4x+2x^{2}\ \textrm{using distributive property}\\\textrm{dividing throughout by two}\\255=2x+x^{2} \\\\x^{2} +2x -255 = 0\\$

Which is a quadratic equation so we will apply splitting the middle term that is factorization we get

∴ $x^{2}+ 17x -15x -255=0\\x(x+17)-15(x+17)=0\\(x-15)(x+17)=0\\\therefore x-15 =0\ or\ x+17=0\\\\\textrm{x cannot be negative i.e -17}\\\therefore x=15$

∴ Width x = 15 yard

∴ Length = 4 + 2x = 4 +2 × 15 = 34 yard

The dimensions of the rectangle are

length = 34 yards

width = 15 yards

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2 years ago

Which of the following representations are functions?

Hitman42 [59]

The answer to this question is a and c only

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2 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

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ch4aika [34]

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Step-by-step explanation:

If you reflect it over the y-axis the first number will change to a positive and the second number will stay the same.

Hope this helps have a great day! :)

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likoan [24]

The number is 40243 :)
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