All categories

3 years ago

If points F and G are contained in a plane, then <-FG-> is entirely contained in that plane

Mathematics

1 answer:

Mkey [24]3 years ago

4 0

The answer is, True.

You might be interested in

john works at a grocery store and makes $11 amount of dollars an hour. how much will be in johns paycheck this week if he worked

Arlecino [84]

Answer: $352

Step-by-step explanation:

6 0

2 years ago

A parking garage charges $1.50 for the 1st hour of parking and $0.75 for each additional hour. What is the cost in dollars for p

QveST [7]

The equation you have to solve is 1.50 + (0.75 • 6). First, you solve what is in the parenthesis, 0.75 • 6 = 4.50. Now you simply add 1.50 and 4.50 to get $6, the total cost for 7 hours worth of parking.

8 0

3 years ago

In 3 apples cost 48¢ , how many dozen apples can be bougth for $3.84?

inn [45]

Ok so let’s do .48/3 then divide 3.84 by our answer to get 24

7 0

3 years ago

Read 2 more answers

What is the recursive rule for the following sequence: -5, -3, -1, 1…

Irina18 [472]

$a_{n}$ = $a_{n-1}$ + 2

So C.

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