Can someone plz help me with all these problems

grandymaker [24]

The answer is: [C]: " f(c) = $\frac{9}{5}$ c + 32 " .
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Explanation:
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Given the original function:

" c(y) = (5/9) (x − 32) " ; in which "x = f" ; and "y = c(f) " ;
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→ 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:
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x = (5/9) (y − 32) ;

Now, rewrite THIS equation; by solving for "y" ; in terms of "x" ;
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→ 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 .
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As such:
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" ( $\frac{5}{9}$ ) * (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:
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→ 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" ;
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So, the inverse function is: " f(c) = (9/5) c + 32 " .
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The answer is: " f(c) = $\frac{9}{5}$ c + 32 " ;
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→ which is:

→ Answer choice: [C]: " f(c) = $\frac{9}{5}$ c + 32 " .
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6 0

2 years ago

What is the answer to the equation (x-5)^2=40

Citrus2011 [14]

Answer:

x^2 -10x - 15 = 0

(don't know how far you want me to go)

Step-by-step explanation:

(x - 5)^2 = 40

(expand the brackets)

(x - 5)(x - 5) = 40

x^2 -5x -5x + 25 = 40

x^2 -10x + 25 = 40

x^2 -10x - 15 = 0

3 0

2 years ago

Relate-to-a-Simpler-Problem

Alex73 [517]

Answer:

1. 128 pages.

2. (y - x) + 1

3. 6 pages.

4. Helps to formulate the expression for the number of pages read.

Step-by-step explanation:

1. George will read pages 385 through 512 in his karate history book. And we are asked to find the number of pages that he will read.

The required number of pages that George read will be = (512 - 385) + 1 = 128 pages.

2. If he read from page number x through page number y then the general rule of finding the number pages that he read will be = (y - x) + 1

3. If x = 500 and y = 505, then the number of pages read is = (505 - 500) + 1 = 6 pages.

Again, if x = 5 and y = 29, then again the number of pages read is = (29 - 5) + 1 = 25 pages.

4. The rule just formulates the expression for the number of pages read and it eases the calculation for the number of pages read. (Answer)

7 0

2 years ago

1. What is the area of this polygon?

siniylev [52]

Answer:

Part 1) The area of the polygon is $42\ units^{2}$

Part 2) The area of the polygon is $43.5\ units^{2}$

Step-by-step explanation:

Part 1) we know that

The area of the figure is equal to the area of rectangle GPST plus the area of triangle PES

Observing the graph

Find the area of rectangle

$A=bh=5*7=35\ units^{2}$

Find the area of triangle

$A=(1/2)bh=(1/2)*7*2=7\ units^{2}$

The area of the figure is equal to

$35\ units^{2}+7\ units^{2}=42\ units^{2}$

see the attached figure N 1 to better understand the problem

Part 2) we know that

The area of the figure is equal to the area of two triangles plus the area of a trapezoid

Observing the graph

Find the area of triangle 1

$A=(1/2)bh=(1/2)*3*2=3\ units^{2}$

Find the area of triangle 2

$A=(1/2)bh=(1/2)*5*1=2.5\ units^{2}$

Find the area of trapezoid

$A=(1/2)(b1+b2)h=(1/2)*(11+8)*4=38\ units^{2}$

The area of the figure is equal to

$3\ units^{2}+2.5\ units^{2}+38\ units^{2}=43.5\ units^{2}$

see the attached figure N 2 to better understand the problem

3 0

2 years ago

Select the correct answer.

postnew [5]

Answer:

About 80% of the students would prefer the shorter school day

Step-by-step explanation:

20/25=.8

80/100=.8

If the results are similar then that will be your answer

8 0

2 years ago