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3 years ago

The total of 42, 75, 189, 301, and 728 is

2 answers:

7 0

The total means add so,
42 + 75 + 189 + 301 + 728 = 117 + 189 + 301 + 728 = 306 + 301 + 728 = 607<span> + 728 </span>=1,335

<span>So, the answer is </span>C.

nordsb [41]3 years ago

5 0

The total means add so,
42 + 75 + 189 + 301 + 728 = 1,335
Answer is C.

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If f(x)=(x+6)(x−3)/(x−2), then calculate f(5). Round answers to two decimal places.

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Step-by-step explanation:

f(5)=(5+6)(5-3)/(5-2)

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4 years ago

Read 2 more answers

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':

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

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

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