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3 years ago

What does x equal 6x+3=x+8

Mathematics

2 answers:

Brrunno [24]3 years ago

5 0

It is 1.................

babunello [35]3 years ago

3 0

X =1. 6+3=9 =1+8 ( =9)

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The ellipse with x-intercepts (2, 0) and (-2, 0); y-intercepts (0, 4) and (0, -4).

masya89 [10]

Answer:

$\dfrac{x^2}{4}+\dfrac{y^2}{16}=1$

Step-by-step explanation:

If the ellipse has its x-intercepts at points (2, 0) and (-2, 0) and y-intercepts at points (0, 4) and (0, -4), then its symmetric across the y-axis and across the x-axis.

Moreover,

$a=2\\ \\b=4$

The equation of such ellipse is

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

Hence, the equation of the ellipse is

$\dfrac{x^2}{2^2}+\dfrac{y^2}{4^2}=1\\ \\ \\\dfrac{x^2}{4}+\dfrac{y^2}{16}=1$

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3 years ago

The data set represents the total number of tickets each person purchased for a play

expeople1 [14]

Answer: would be 10

Step-by-step explanation:

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3 years ago

Which of the following expressions is equal to 1 divided by 16

Mama L [17]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

9(x + y) = 9x + 9y what property is that

kogti [31]

The distributive property

4 0

3 years ago

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

3 years ago