If AB = 2x, and BC = x+5 and AC = 17 find X

How many

Westkost [7]

Answer:

There are no solutions

Step-by-step explanation:

4/5 ( 20x + 5 ) = 16x - 2

Distibute on the left:

( 4/5 x 20 x ) + ( 4/5 x 5 ) = 16x - 2

16x + 4 = 16x - 2

Subtract 16x from both sides:

4=-2

Add 2 to each side:

6=0

No solutions

8 0

2 years ago

Read 2 more answers

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

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

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

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

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

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

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

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

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

3 years ago

Read 2 more answers

Name the property of equality that justifies: (I have to put question into an equation below because I can't get it in this ques

Rainbow [258]

The best and most correct answer among the choices provided by your question is the third choice or letter C "Multiplication Property of Equality"

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

7 0

3 years ago

In MON, J, K, and L are midpoints. If JL = 11, LK = 13, and ON = 20, and JL || MN, LK || MO, and JK || ON, what is the length of

In-s [12.5K]

Answer:

The lengths of MN is 22 units, MO is 26 units and JK is 10 units

Step-by-step explanation:

A line segment joining the mid-points of two sides in a triangle is parallel to the third side and equal to half its length

In Δ MON

∵ J, K, and L are mid-points

∵ JL // MN and LK // MO

∴ L is the mid-point of ON

∴ J is the mid-point of MO

∴ K is the mid-point of MN

∵ J, L are the mid-points of MO and ON

∵ JL is opposite to MN

→ By using the rule above

∴ JL = $\frac{1}{2}$ MN

∵ JL = 11 units

∴ 11 = $\frac{1}{2}$ MN

→ Multiply both sides by 2

∴ 22 = MN

∴ MN = 22 units

∵ K, L are the mid-points of MN and ON

∵ KL is opposite to MO

→ By using the rule above

∴ KL = $\frac{1}{2}$ MO

∵ KL = 13 units

∴ 13 = $\frac{1}{2}$ MO

→ Multiply both sides by 2

∴ 26 = MO

∴ MO = 26 units

∵ J, K are the mid-points of MO and MN

∵ JK is opposite to ON

→ By using the rule above

∴ JK = $\frac{1}{2}$ ON

∵ ON =20 units

∴ JK = $\frac{1}{2}$ (20)

∴ JK = 10 units

∴ The lengths of MN are 22 units, MO is 26 units and JK is 10 units

3 0

3 years ago

Yes or no awnser pls

kifflom [539]

Answer:

No the question iant a function

5 0

2 years ago