Someone please help im stuck

Mathematics

1 answer:

BARSIC [14]3 years ago

4 0

The answer is b, -11.
plug in 3 as the input and solve.

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Noah wants to buy a purebed golden retriever puppy that he sees for $499.By the time he goes to buy it, the price has increased

eduard

499 - 100%
599 - x

x = 599*100/499 = 120.04
x-100% = 120.04-100 = 20.04

Answer: The price increased by 20%.

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a state park charges $5 per car plus $1 per person as an admission fee. the total charged for a car with x people is f(x)

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Answer:

the total charged for a car is f(x)=x+5

Step-by-step explanation:

because ther is only on car the value 5 could be constant. but because there could be a number of people in the car that is described by x and is multiplied by the cost which is 1

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The line segment AB with endpoints A (-3, 6) and B (9, 12) is dilated with a scale

Gwar [14]

Answer:

C) (-2, 4), (6,8) is the correct answer.

Step-by-step explanation:

Given that line segment AB:

A (-3, 6) and B (9, 12) is dilated with a scale factor 2/3 about the origin.

First of all, let us calculate the distance AB using the distance formula:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here,

$x_2=9\\x_1=-3\\y_2=12\\y_1=6$

Putting all the values and finding AB:

$AB = \sqrt{(9-(-3))^2+(12-6)^2}\\\Rightarrow AB = \sqrt{(12)^2+(6)^2}\\\Rightarrow AB = \sqrt{144+36}\\\Rightarrow AB = \sqrt{180}\\\Rightarrow AB = 6\sqrt{5}\ units$

It is given that AB is dilated with a scale factor of $\frac{2}{3}$ .

$x_2'=\dfrac{2}{3}\times x_2=\dfrac{2}{3}\times9=6\\x_1'=\dfrac{2}{3}\times x_1=\dfrac{2}{3}\times-3=-2\\y_2'=\dfrac{2}{3}\times y_2=\dfrac{2}{3}\times 12=8\\y_1'=\dfrac{2}{3}\times y_1=\dfrac{2}{3}\times 6=4$

So, the new coordinates are A'(-2,4) and B'(6,8).

Verifying this by calculating the distance A'B':

$A'B' = \sqrt{(6-(-2))^2+(8-4)^2}\\\Rightarrow A'B' = \sqrt{(8)^2+(4)^2}\\\Rightarrow A'B' = \sqrt{64+16}\\\Rightarrow A'B' = \sqrt{80}\\\Rightarrow A'B' = 4\sqrt{5}\ units = \dfrac{2}{3}\times AB$