Solve for m. 13m−23=113

attashe74 [19]

This would be easiest to do by completing the square.

x^2+10x-2=0

x^2+10x=2

x^2+10x+25=27

(x+5)^2=27

x+5=±√27

x=-5±√27

(x+5+√27)(x+5-√27)

5 0

3 years ago

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Create a sequence with five terms based on the given information. a. A arithmetic sequence with a third term of 8 and a constant

kolezko [41]

Part a .

A arithmetic sequence with a third term of 8 and a common difference of 5 .

To find the first five therms, since the common difference is 5, so we add 5 to get the fourth term and add 5 to fourth term to get the fifth term .

And for first two terms, we will subtract 5 from 8 to get the second term and subtract 5 from the second term to get the first term. And we will get

$8-5-5,8-5,8, 8+5+5 = -2 ,3,8,13,18$

Part b:A geometric sequence with a fifth term of 1/3 and constant ratio of 1/3.

TO find the first five terms, since the constant ratio of 1/3, so we multiply 1/3 to third term to get fourth term, and multiply fourth term by 1/3 to get fifth term .

And to get the first two terms, we will divide third term by 1/3, to get the second term and divide the second term by 1/3 to get the first term, that is

$3,1,\frac{1}{3}, \frac{1}{9},\frac{1}{27}$

6 0

3 years ago

Write a minimum 100-word essay on how you can currently use and may use your knowledge and skills of rotations and it's use in d

Delvig [45]

Did you really think someone would write you a whole essay just for fun

3 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

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Which graph shows the solution to the equation below? log Subscript 3 Baseline (x + 3) = log Subscript 0.3 (x minus 1) On a coor

Effectus [21]

Answer:

On a coordinate plane, 2 curves intersect at (1, 1). One curve curves up and to the right from quadrant 3 into quadrant 1. The other curve curves down from quadrant 1 into quadrant 4

Step-by-step explanation:

The first function is given as:

$log_3(x+3)$