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2 years ago

Which of the following is a sequence?

Mathematics

1 answer:

Anna71 [15]2 years ago

6 0

Answer:

The answer of this question is D

Step-by-step explanation:

Time 2

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If the two lines below are perpendicular and the slope of the red line is -7,

Tems11 [23]

Answer:

C. ⅐

Step-by-step explanation:

Recall: the slope of a line that is perpendicular to another is the negative reciprocal of the slope of the other line that it is perpendicular to.

Thus:

Slope of red line = -7

The green line that is perpendicular to the red line will have a slope that is the negative reciprocal of -7.

Negative reciprocal of -7 = ⅐

The slope of the green line is therefore ⅐

5 0

2 years ago

What is hte vertex of f(x) = 5x^+20x-16

hjlf

What is the exponent ?

6 0

3 years ago

Best buy sells iPhones amd Samsung. Thy have 81 phones in the store. If the number of iphones is twice the number of Samsung's h

seropon [69]

Answer:

54 iphones ,27 samsungs

81/3=27

27x2=54

81-54=27

5 0

2 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

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

URGENT EXPLANATION AND SOLVING

Alex787 [66]

$\cfrac{a}{\sin(a)} = \cfrac{b}{\sin(b)} 
$

$\cfrac{31}{\sin(42)} = \cfrac{37}{\sin(x)} 
$

$31\sin(x) = 37 \sin(42)$

$\sin(x) = \cfrac{37 \sin(42) }{31}$

$x = \sin^{-1}(\cfrac{37 \sin(42) }{31} )$

$x = 53 \textdegree$

Answer: 53°

7 0

3 years ago

Read 2 more answers