59÷2 <img src="https://tex.z-dn.net/?f=59%20%5Cdiv%202" id="TexFormula1" title="59 \div 2" alt="59 \div 2" align="absmiddle"

Find the value of x PLEASE EXPLAIN

Wewaii [24]

Answer:

x = 57, x = 111

Step-by-step explanation:

The sum of the 3 angles in a triangle = 180°

Sum the 3 angles and equate to 180, that is

x + 67 + 56 = 180

x + 123 = 180 ( subtract 123 from both sides )

x = 57

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Similarly

x + 47 + 22 = 180

x + 69 = 180 ( subtract 69 from both sides )

x = 111

6 0

2 years ago

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23 x 2 EZ question IK I’m just stoopid

mash [69]

It would be 46, have a nice day

3 0

2 years ago

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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

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(5 + 4 - 2) x (-2) = ? A. -22 B. 22 C. 14 D. -15

brilliants [131]

It would actually be -14 because 5+4-2=7 and if you multiply 7 by -2 you get -14

4 0

3 years ago

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3 = –(–y + 6) a. –9 b. –3 c. 3 d. 9

Vlad1618 [11]

3 = -(-y + 6)

First, simplify brackets. / Your problem should look like: 3 = y - 6
Second, add 6 to both sides. / Your problem should look like: 3 + 6 = y
Third, simplify 3 + 6 to 9. / Your problem should look like: 9 = y
Fourth, switch sides. / Your problem should look like: y = 9

The answer is D) 9.

7 0

3 years ago