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3 years ago

PLEASE ANSWER EACH OF THESE QUESTIONS IM GIVING OUT BRAINLIEST !!!!! D:

Mathematics

2 answers:

Alexus [3.1K]3 years ago

3 0

Answer:

no 1

2x^2 + 6

no 2

x^2 + 63

no3

2x^2 + 20

no 4

3 x 3 x 4 divided by 3

=12cm^3

kompoz [17]3 years ago

3 0

Answer:

hypotenuse=15 ft

Step-by-step explanation:

(2x+1)(x+6)=2x^2+13x+6

(x+7)(x+9)=x^2+16x+63

(2x+4)(x+5)=2x^2+14x+20

4 no question

V=L*B*H/3

=3*3*4/3

=12 in^3

L=lenghth

B=breadth

H=height

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Choose the domain for which each function is defined

klemol [59]

Answer:

Part 1) $f(x)=\frac{x+4}{x}$ -----> $x\neq 0$

Part 2) $f(x)=\frac{x}{x+4}$ ----> $x\neq -4$

Part 3) $f(x)=x(x+4)$ ----> All real numbers

Part 4) $f(x)=\frac{4}{x^2+8x+16}$ ----> $x\neq -4$

Step-by-step explanation:

we know that

The domain of a function is the set of all possible values of x

Part 1) we have

$f(x)=\frac{x+4}{x}$

we know that

In a quotient the denominator cannot be equal to zero

For the value of x=0 the function is not defined

therefore

The domain is

$x\neq 0$

Part 2) we have

$f(x)=\frac{x}{x+4}$

we know that

In a quotient the denominator cannot be equal to zero

For the value of x=-4 the function is not defined

therefore

The domain is

$x\neq -4$

Part 3) we have

$f(x)=x(x+4)$

Applying the distributive property

$f*(x)=x^2+4x$

This is a vertical parabola open upward

The function is defined by all the values of x

therefore

The domain is all real numbers

Part 4) we have

$f(x)=\frac{4}{x^2+8x+16}$

we know that

In a quotient the denominator cannot be equal to zero

Equate the denominator to zero

$x^2+8x+16=0$

Remember that

$x^2+8x+16=(x+4)^2$

( $x+4)^2=0$

The solution is x=-4

For the value of x=-4 the function is not defined

therefore

The domain is

$x\neq -4$

6 0

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Please help will give brainliest

Olegator [25]

Answer:

I hope u understand ...

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