Pls help ASAP!I will mark u brainliest, and pls answer correctly or I will report u A printing company is enlarging the image on a postcard to make a greeting card. The enlargement of the postcard’s rectangular image is done using a scale factor of 125%. Be sure to show all other related math work used to answer the following questions.
(a)Represent a scale factor of 125% as a fraction and decimal.
(b)The postcard’s dimensions are 7 inches by 5 inches. What are the dimensions of the greeting card?
(c)If the printing company makes a poster by enlarging the postcard image, and the poster’s dimensions are 28 inches by 20 inches, represent the scale factor as a percent.
(d)Write an equation, in terms of the scale factor, that shows the relationship between the areas of the
postcard and poster. Explain your equation.
(e)Suppose the printing company wanted to start with the greeting card’s image and reduce it to create the postcard’s image. What scale factor would they use? Represent this scale factor as a percent.
(f). In math class, students had to create a scale drawing that was smaller than the postcard image. Azra
used a scale factor of 60% to create the smaller image. She stated the dimensions of her smaller image as 4 1/6 inches by 3 inches. Azra’s math teacher did not give her full credit for her answer. Why? Explain Azra’s error, and write the answer correctly

Step-by-step explanation:

homozygous. one is homozygous, while two is heterozygous.

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).