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3 years ago

11

A model airplane has a length of 15in and a width of 9in. If that actual airplane is going to be 65 feet long, how wide is the a

ctual plane?
A. 42ft
B. 58ft
C. 39ft
D. 23ft

1 answer:

Vaselesa [24]3 years ago

3 0

The answer would be C.

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Mrrafil [7]

$5=x^2+2x\\
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x^2+2x+1-6=0\\
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3 0

3 years ago

Carlos is a boat dealer who bought a sailor for 11,400 and sold it for 12,711 what percent was his markup on a boat

dangina [55]

So we have 11400 +11400×x=12711
this represents the original price plus some percentage of the original price is equal to the new price. now all we need to do is solve for x
so 11400+11400x=12711
subtract 11400 from each side
11400x=1311 now divide both sides by 11400
x=.115 or about 12%
we can. check by finding 12% of the original price.
11400×.12=1368
now add to 11400
11400+1368=12768 so we are a bit over. this is because it is actually 11.5% just depends on how accurate your answer needs to be

6 0

3 years ago

Determine what type of model best fits the given situation:

lyudmila [28]

Let value intially be = P

Then it is decreased by 20 %.

So 20% of P = $\frac{20}{100} \times P = 0.2P$

So after 1 year value is decreased by 0.2P

so value after 1 year will be = P - 0.2P (as its decreased so we will subtract 0.2P from original value P) = 0.8P-------------------------------------(1)

Similarly for 2nd year, this value 0.8P will again be decreased by 20 %

so 20% of 0.8P = $\frac{20}{100} \times 0.8P = (0.2)(0.8P)$

So after 2 years value is decreased by (0.2)(0.8P)

so value after 2 years will be = 0.8P - 0.2(0.8P)

taking 0.8P common out we get 0.8P(1-0.2)

= 0.8P(0.8)

$=P(0.8)^{2}$ -------------------------(2)

Similarly after 3 years, this value $P(0.8)^{2}$ will again be decreased by 20 %

so 20% of $P(0.8)^{2} \frac{20}{100} \times P(0.8)^{2} = (0.2)P(0.8)^{2}$

So after 3 years value is decreased by $(0.2)P(0.8)^{2}$

so value after 3 years will be = $P(0.8)^{2} - (0.2)P(0.8)^{2}$

taking $P(0.8)^{2}$ common out we get $P(0.8)^{2}(1-0.2)$

$P(0.8)^{2}(0.8)$

$P(0.8)^{3}$ -----------------------(3)

so from (1), (2), (3) we can see the following pattern

value after 1 year is P(0.8) or $P(0.8)^{1}$

value after 2 years is $P(0.8)^{2}$

value after 3 years is $P(0.8)^{3}$

so value after x years will be $P(0.8)^{x}$ ( whatever is the year, that is raised to power on 0.8)

So function is best described by exponential model

$y = P(0.8)^{x}$ where y is the value after x years

so thats the final answer

3 0

3 years ago

In what quadrant is the terminal side<br> of -323°?

dexar [7]

Answer:

Quadrant 4 i believe plz tell me if im wrong

Step-by-step explanatio

6 0

3 years ago

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What is 2.25 of 8 PLEASE HELP!!!!!

Pani-rosa [81]

The answer is 18 because u multiply when u see the word of

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3 years ago

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