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3 years ago

If a(x)=2x-4 and b(x) =x+2, which of the following expressions produces a quadratic function?

Mathematics

1 answer:

ZanzabumX [31]3 years ago

5 0

Answer:

$(ab)(x)$

Step-by-step explanation:

The complete question in the attached figure

we have

$a(x)=2x-4\\b(x)=x+2$

we know that

(ab)(x) is equal to the product of a(x) and b(x)

$(ab)(x)=(2x-4)(x+2)$

Applying distributive property

$(ab)(x)=2x(x)+2x(2)-4(x)-4(2)$

$(ab)(x)=2x^2+4x-4x-8$

$(ab)(x)=2x^2-8$

This is a quadratic equation

The other expressions not produce a quadratic equation

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What is the sum of all integers between 19 and 77 squared rooted?

Alex Ar [27]

4<√19<5
8<√77<9
so the quality numbers are 5,6,7,and 8
5+6+7+8=26
so the answer is 26

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3 years ago

I have nooooo clue, please help

Aleksandr-060686 [28]

Answer:

case a) $x^{2}=3y$ ----> open up

case b) $x^{2}=-10y$ ----> open down

case c) $y^{2}=-2x$ ----> open left

case d) $y^{2}=6x$ ----> open right

Step-by-step explanation:

we know that

1) The general equation of a vertical parabola is equal to

$y=a(x-h)^{2}+k$

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open upward and the vertex is a minimum

If a<0 ----> the parabola open downward and the vertex is a maximum

2) The general equation of a horizontal parabola is equal to

$x=a(y-k)^{2}+h$

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open to the right

If a<0 ----> the parabola open to the left

Verify each case

case a) we have

$x^{2}=3y$

$y=(1/3)x^{2}$

$a=(1/3)$

$a>0$

therefore

The parabola open up

case b) we have

$x^{2}=-10y$

$y=-(1/10)x^{2}$

$a=-(1/10)$

$a$

therefore

The parabola open down

case c) we have

$y^{2}=-2x$

$x=-(1/2)y^{2}$

$a=-(1/2)$

$a$

therefore

The parabola open to the left

case d) we have

$y^{2}=6x$

$x=(1/6)y^{2}$

$a=(1/6)$

$a>0$

therefore

The parabola open to the right

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Answer:

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Step-by-step explanation:

1hr=$12.4

42hrs=x

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Cloud [144]

Answer:

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Step-by-step explanation:

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