2 years ago

What is the range of the function, f(x)?

Mathematics

1 answer:

BartSMP [9]2 years ago

5 0

Answer:

Step-by-step explanation:

All the y values for f(x) are less than -1 and go upto infinity.

You might be interested in

if a square + b square + c square minus ab minus bc minus ca equals to zero, prove that A equals to B equal to C

lawyer [7]

A^2 + b^2 + c^2 - ab - bc - ca = 0
a.a + b.b + c.c = a.b + b.c + c.a

equating term by term:-
a = b
b = c
c = a

Therefore a = b = c

8 0

3 years ago

5.) What is the GCF of the following<br> 6x2 and 12x3

algol13

Answer: 12
6x2 is 12 and 12x3 is 38.
We will first find the prime factorization of 12 and 36. After we will calculate the factors of 12 and 36 and find the biggest common factor number .

3 0

2 years ago

Find the missing fraction. What is the ??? equal to? (the fractions aren't in order)

dimaraw [331]

Step-by-step explanation:

1/6 divided by 3=1/2.

I think this is the answer

5 0

2 years ago

Solve this x²+8/10 -x = 10/2

AysviL [449]

$x^2+\frac{8}{10}-x=\frac{10}{2}\\
x^2-x+\frac{8}{10}-\frac{10}{2}=0\\
x^2-x+\frac{8}{10}-\frac{50}{10}=0\\
x^2-x-\frac{42}{10}=0\\
x^2-x+\frac{1}{4}-\frac{1}{4}-\frac{42}{10}=0\\
(x-\frac{1}{2})^2=\frac{1}{4}+\frac{42}{10}\\
(x-\frac{1}{2})^2=\frac{5}{20}+\frac{84}{20}\\
(x-\frac{1}{2})^2=\frac{89}{20}\\
x-\frac{1}{2}=\sqrt{\frac{89}{20}} \vee x-\frac{1}{2}=-\sqrt{\frac{89}{20}}\\
x=\frac{1}{2}+\frac{\sqrt{89}}{\sqrt{20}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{\sqrt{20}}\\


$
$x=\frac{1}{2}+\frac{\sqrt{89}}{2\sqrt{5}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{2\sqrt5}}\\
x=\frac{1}{2}+\frac{\sqrt{445}}{10} \vee x=\frac{1}{2}-\frac{\sqrt{445}}{10}}\\
x=\frac{5+\sqrt{445}}{10} \vee x=\frac{5-\sqrt{445}}{10}}\\$

7 0

3 years ago

Mary has 28 Apple she gave her mother 2355 how many apples does he have

stiks02 [169]

Answer:

mary broke

Step-by-step explanation:

it simple she will need to pay 2338 apples

4 0

2 years ago

Read 2 more answers