The pair of figures is similar. Find x. Round to the nearest tenth if necessary.

grandymaker [24]

The answer is: [C]: " f(c) = $\frac{9}{5}$ c + 32 " .
________________________________________________________

Explanation:
________________________________________________________
Given the original function:

" c(y) = (5/9) (x − 32) " ; in which "x = f" ; and "y = c(f) " ;
________________________________________________________
→ Write the original function as: " y = (5/9) (x − 32) " ;

Now, change the "y" to an "x" ; and the "x" to a "y"; and rewrite; as follows:
________________________________________________________
x = (5/9) (y − 32) ;

Now, rewrite THIS equation; by solving for "y" ; in terms of "x" ;
_____________________________________________________
→ That is, solve this equation for "y" ; with "c" as an "isolated variable" on the
"left-hand side" of the equation:

We have:

→ x = " ( $\frac{5}{9}$ ) * (y − 32) " ;

Let us simplify the "right-hand side" of the equation:
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Note the "distributive property" of multiplication:
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a(b + c) = ab + ac ; AND:

a(b – c) = ab – ac .
__________________________________________

As such:
__________________________________________

" ( $\frac{5}{9}$ ) * (y − 32) " ;

= [ ( $\frac{5}{9}$ ) * y ] − [ ( $\frac{5}{9}$ ) * (32) ] ;

= [ ( $\frac{5}{9}$ ) y ] − [ ( $\frac{5}{9}$ ) * ( $\frac{32}{1})$ " ;

= [ ( $\frac{5}{9}$ ) y ] − [ ( $\frac{(5*32)}{(9*1)}$ ] ;

= [ ( $\frac{5}{9}$ ) y ] − [ ( $\frac{(160)}{(9)}$ ] ;

= [ ( $\frac{5y}{9}$ ) ] − [ ( $\frac{(160)}{(9)}$ ] ;

= [ $\frac{(5y-160)}{9}$ ] ;
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And rewrite as:

→ " x = $\frac{(5y-160)}{9}$ " ;

We want to rewrite this; solving for "y"; with "y" isolated as a "single variable" on the "left-hand side" of the equation ;

We have:

→ " x = $\frac{(5y-160)}{9}$ " ;

↔ " $\frac{(5y-160)}{9}$ = x ;

Multiply both sides of the equation by "9" ;

9 * $\frac{(5y-160)}{9}$ = x * 9 ;

to get:

→ 5y − 160 = 9x ;

Now, add "160" to each side of the equation; as follows:
_______________________________________________________

→ 5y − 160 + 160 = 9x + 160 ;

to get:

→ 5y = 9x + 160 ;

Now, divided Each side of the equation by "5" ;
to isolate "y" on one side of the equation; & to solve for "y" ;

→ 5y / 5 = (9y + 160) / 5 ;

to get:

→ y = (9/5)x + (160/5) ;

→ y = (9/5)x + 32 ;

→ Now, remember we had substituted: "y" for "c(f)" ;

Now that we have the "equation for the inverse" ;
→ which is: " (9/5)x + 32" ;

Remember that for the original ("non-inverse" equation); "y" was used in place of "c(f)" . We have the "inverse equation"; so we can denote this "inverse function" ; that is, the "inverse" of "c(f)" as: "f(c)" .

Note that "x = c" ;
_____________________________________________________
So, the inverse function is: " f(c) = (9/5) c + 32 " .
_____________________________________________________

The answer is: " f(c) = $\frac{9}{5}$ c + 32 " ;
_____________________________________________________
→ which is:

→ Answer choice: [C]: " f(c) = $\frac{9}{5}$ c + 32 " .
_____________________________________________________

6 0

3 years ago

Sorry, the last one wasn’t showing the picture. Is this a function or not? Also tell me how you got the answer. TwT ty

kap26 [50]

Answer:

yes this is a function

Step-by-step explanation:

simple method to find if something is a function

if you can draw a vertical line at any point on the graph and get two intersections, it is not a function

7 0

3 years ago

Estimate the student's walking pace, in steps per minute, at 3:20 p.m. by averaging the slopes of two secant lines from part (a)

ehidna [41]

This question is incomplete, the complete question is;

A student bought a smart-watch that tracks the number of steps she walks throughout the day. The table shows the number of steps recorded (t) minutes after 3:00 pm on the first day she wore the watch.

t (min) 0 10 20 30 40

Steps 3,288 4,659 5,522 6,686 7,128

a) Find the slopes of the secant lines corresponding to the given intervals of t.

1) [ 0, 40 ]

11) [ 10, 20 ]

111) [ 20, 30 ]

b) Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part (a). (Round your answer to the nearest integer.)

Answer:

a)

1) for [ 0, 40 ], slope is 96

11) for [ 10, 20 ], slope is 86.3

111) for [ 20, 30 ], slope is 116.4

b) the student's walking pace is 101 per min

Step-by-step explanation:

Given the data in the question;

t (min) 0 10 20 30 40

Steps 3,288 4,659 5,522 6,686 7,128

SLOPE OF SECANT LINES

1) [ 0, 40 ]

slope = ( 7,128 - 3,288 ) / ( 40 - 0

= 3840 / 40 = 96

Hence slope is 96

11) [ 10, 20 ]

slope = ( 5,522 - 4,659 ) / ( 20 - 10 )

= 863 / 10 = 86.3

Hence slope is 86.3

111) [ 20, 30 ]

slope = ( 6,686 - 5,522 ) / ( 30 - 20 )

= 1164 / 10 = 116.4

Hence slope is 116.4

b)

Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part .

Since this is recorded after 3:00 pm

{ 3:20 - 3:00 = 20 }

so t = 20 min

so by average;

we have ( [ 10, 20 ] + [ 20, 30 ] ) /2

⇒ ( 86.3 + 116.4 ) / 2

= 202.7 /2

= 101.35 ≈ 101

Therefore, the student's walking pace is 101 per minutes

3 0

3 years ago

Reflect the given preimage over y=−1 followed by y=−7. Find the new coordinates. What one transformation is this double reflecti

fenix001 [56]

Answer:

Reflecting over y = -1 line:

A'(8, -10)

B'(10, -8)

C'(2, -4)

Reflecting over y = -7 line:

A''(8, -4)

B'(10, -6)

C''(2, -10)

Step-by-step explanation:

Reflect the given preimage over y=−1 followed

by y=−7. Find the new coordinates. What one transformation is this double reflection the same as? (Note: when you are reflecting over a y= line, the x-values of the preimage will remain the same and you will be changing the y-values)

The coordinates of the preimage are:

A(8,8)

B(10,6)

C(2,2)

Answer: Reflecting over y = -1:

If a point is reflected over a y line, the x values remain the same while the y values change.

For point A(8, 8): The y distance between the y = - 1 line and point A is 9 units. (8- (-1)). If point A is reflected, the y value would be 9 units below the y = -1 line, i.e the new y coordinate would be at -10 (-1-9)

The new coordinate is at A'(8, -10)

For point B(10, 6): The y distance between the y = - 1 line and point B is 7 units. (6- (-1)). If point B is reflected, the y value would be 7 units below the y = -1 line, i.e the new y coordinate would be at -8 (-1-7)

The new coordinate is at B'(10, -8)

For point C(2, 2): The y distance between the y = - 1 line and point C is 3 units. (2- (-1)). If point C is reflected, the y value would be 3 units below the y = -1 line, i.e the new y coordinate would be at -4 (-1-3)

The new coordinate is at C'(2, -4)

Reflecting over y = -7 line:

For point A'(8, -10): The y distance between the y = - 7 line and point A' is 3 units. (-7- (-10)). If point A' is reflected, the y value would be 3 units above the y = -7 line, i.e the new y coordinate would be at -4 (-7+3)

The new coordinate is at A''(8, -4)

For point B'(10, -8): The y distance between the y = - 7 line and point B' is 1 units. (-7- (-8)). If point B' is reflected, the y value would be 1 units above the y = -7 line, i.e the new y coordinate would be at -6 (-7 + 1)

The new coordinate is at B'(10, -6)

For point C'(2, -4): The y distance between the y = - 7 line and point C' is 3 units. (-4- (-7)). If point C' is reflected, the y value would be 3 units below the y = -7 line, i.e the new y coordinate would be at -10 (-7-3)

The new coordinate is at C''(2, -10)

We can also see that h= −7−(−1)=−6. We know that two reflections is the same as a translation of 2h units down. So 2(−6) is a translation of −12 units down.

5 0

3 years ago

a person invested $6300 for 1 year part at 8% part at 10% and the remainder at 15%. The total annual income from these investmen

Lunna [17]

Answer:

Step-by-step explanation:

let x be part at 8%, y be part at 10% and z be part at 15%

(1) x+y+z=6300

the amount of money invested at 15% was $100 more than the amounts invested at 8% and 10% combined. so

(2) z=x+y+100

The total annual income from these investments was $766. so

0.08x+0.1y+0.15z=766 or

(3) 8x+10y+15z=76600

substitute (2) into (1)

x+y+x+y+100=6300

(4) 2x+2y=6200

substitute (2) inot (3)

8x+10y+15(x+y+100)=76600

(5) 23x+25y=75100

solving (4) n (5)

x=1200

y=1900

z=1200+1900+100=3200

7 0

3 years ago

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