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2 years ago

I need help please!!!

Mathematics

2 answers:

nikdorinn [45]2 years ago

6 0

Answer: 15a+8b

Step-by-step explanation:

lana66690 [7]2 years ago

5 0

Answer:

15a + 8b

you multiply 2 by the numbers in the parantasis so our new equasion will be 10a + 6b +5a +2b. You can only add numbers that have THE SAME VARIABLE. So 10a+5a=15a and 6b+2b=8b so our answer is 15a+8b

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Can someone please help?

Firlakuza [10]

Answer:

1. x+4x

2. 5x

3. $6\frac{7}{x}$

4. 4x-2x+15

5. $(6+x)^2$

Step-by-step explanation:

To write expressions remember these key words:

a number is a variable like x
sum is +
difference and subtracted is -
quotient is ÷
product is ×
reciprocal means flip like $\frac{1}{9}---> \frac{9}{1}$
square is an exponent of 2 like $x^2$

1. x+4x

2. 5x

3. $6\frac{7}{x}$

Remember a quotient would be $\frac{x}{7}$ but the reciprocal is $\frac{7}{x}$ .

4. 4x-2x+15

5. $(6+x)^2$

6 0

3 years ago

Which pair of funtions is not a pair of inverse functions? please help!!

antiseptic1488 [7]

Answer:

$f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Step-by-step explanation:

we know that

To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.

we will proceed to verify each case to determine the solution of the problem

<u>case A)</u> $f(x)=\frac{x+1}{6} , g(x)=6x-1$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y+1}{6}$

Isolate the variable y

$6x=y+1$

$y=6x-1$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=6x-1$

therefore

f(x) and g(x) are inverse functions

<u>case B)</u> $f(x)=\frac{x-4}{19} , g(x)=19x+4$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y-4}{19}$

Isolate the variable y

$19x=y-4$

$y=19x+4$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=19x+4$

therefore

f(x) and g(x) are inverse functions

<u>case C)</u> $f(x)=x^{5}, g(x)=\sqrt[5]{x}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=y^{5}$

Isolate the variable y

fifth root both members

$y=\sqrt[5]{x}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\sqrt[5]{x}$

therefore

f(x) and g(x) are inverse functions

<u>case D)</u> $f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y}{y+20}$

Isolate the variable y

$x(y+20)=y$

$xy+20x=y$

$y-xy=20x$

$y(1-x)=20x$

$y=20x/(1-x)$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=20x/(1-x)$

$\frac{20x}{1-x}\neq \frac{20x}{x-1}$

therefore

f(x) and g(x) is not a pair of inverse functions

7 0

2 years ago

Read 2 more answers

Please help urgent ❤️❤️❤️Pl

olga nikolaevna [1]

Answer:

theres no question so we cant answer nothing

5 0

3 years ago

How many more quarters are in $4.50 than in mexican pasos

natali 33 [55]

Answer:

76.39

Step-by-step explanation:

$4=76.39 paysos

6 0

2 years ago

20 points!! Pls help

Studentka2010 [4]

Answer:

Step-by-step explanation:

-x-3=y, therefore x = -y-3

-3x - 8y = 4

Find the value of y

Substitute the x in -3x - 8y = 4 with -y-3

We get

-3·(-y-3) - 8y = 4

3y + 9 - 8y = 4

-5y = 4-9

-5y = -5

_______________

Find the value of x

x = -y-3

x = -1-3

_______________

Answer (x, y) =

_____________________

#IndonesianPride - kexcvi

5 0

3 years ago

Read 2 more answers