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2 years ago

Please help! im so lost..

Mathematics

1 answer:

Tom [10]2 years ago

7 0

i wish i knew but you got this

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HELP PLEASE FIND X AND SHOW YOUR WORK

PtichkaEL [24]

Answer:

Step-by-step explanation:

Straight lines are 180 degrees, so the angle inside of the triangle next to the 127, is 53 degrees. (180-127= 53)

So to find "x", we need to figure out what the full number is.

59+53 = 112.

180-112= 68.

Now that we have the degree of the full angle, we can make an equation to find x.

x*2 = 68

(Divide 2 from both sides to isolate the variable)

x = 34

I hope this helped!

7 0

3 years ago

Alan had already read 12 books this year before joining a book club, and he plans to read 1

Verizon [17]

Respuesta:

Step-by-step explanation:

7 libros

12 - 19 ⇒ 7

5 0

3 years ago

How many ways can four people (A, B, C, D) sit in a row at a move theater if C and D must sit next to each other?

Novay_Z [31]

Answer:

24 ways in four people.

6 0

3 years ago

Q1 : Find the inverse of the function.

Mamont248 [21]

Answer:

1. Option D is correct.

2. Option B is correct.

3. Option B is correct.

Step-by-step explanation:

Inverse function defined as the the function that undergoes the action of the other function.

A function $f^{-1}$ is the inverse of f if whenever y =f(x) and $x =f^{-1}$

To find the inverse of the function:

Q1.

Given the function: f(x) = 7x -1

Put y for f(x) and solve for x;

y= 7x -1

Add 1 both sides we get;

y + 1 = 7x

Divide both sides by 7 we get;

$x = \frac{y+1}{7}$

Put $f^{-1}(y)$ for x;

$f^{-1}(y) = \frac{y+1}{7}$

Interchange y =x, we have

$f^{-1}(x) = \frac{x+1}{7}$

Q 2.

Given the function:

$f(x) = x^3 - 7$

Put y for f(x) and solve for x;

$y = x^3-7$

Add 7 both sides we get;

$y + 7 =x^3$

taking cube root both sides we get

$x =\sqrt[3]{y+7}$

Put $f^{-1}(y)$ for x;

$f^{-1}(y) =\sqrt[3]{y+7}$

Interchange y =x, we have

$f^{-1}(x) = \sqrt[3]{x+7}$

Q3 .

Given the function:

$f(x) = 5x^3 - 3$

Put y for f(x) and solve for x;

$y = 5x^3-3$

Add 3 both sides we get;

$y + 3 =5x^3$

Divide both sides by 5 we get;

$x^3 = \frac{y+3}{5}$

taking cube root both sides we get

$x = \sqrt[3]{\frac{y+3}{5} }$

Put $f^{-1}(y)$ for x;

$f^{-1}(y) = \sqrt[3]{\frac{y+3}{5} }$

Interchange y =x, we have

$f^{-1}(x) = \sqrt[3]{\frac{x+3}{5} }$

7 0

3 years ago

Marcus is trying to find 4 5/6-1 3/6. His work shown. What is Marcus's mistake?

mamaluj [8]

Answer:

Step 3

Step-by-step explanation:

The mistake was made in Step 3.

Step 1: Subtract the wholes. 4 - 1 = 3

Step 2: Subtract the fractions. 5/6 - 3/6 = 2/6

After Step 2, He should have added them instead of subtracting them:

3 + 2/6 = 3 2/6

So, step 3 was his mistake.

6 0

3 years ago

Please help! im so lost..​

Please help! im so lost..