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adelina 88 [10]
3 years ago
10

Linda and Roy are solving the equation 4x+3x-7=(1+4x)+5x. Linda uses a first step that results in 4x+7-3x=(1+4x)+5x Roy uses a f

irst step that results in 4x+3x-7=1+(4x+5x) which statement about the first steps Linda and Roy use is true

Mathematics
1 answer:
White raven [17]3 years ago
4 0

Answer:

Option C

Step-by-step explanation:

Linda used the first step that results in,

4x + 7 - 3x = (1 + 4x) + 5x

As Linda changed the sign of (3x) to (-3x), it doesn't matches the original equation.

So she is not correct.

Roy used the first step as,

4x + 3x - 7 = 1 + (4x + 5x)

He has used the associative property in the right hand side of the given equation.

Associative property → (a + b) + c → a + (b + c)

Therefore, he is correct.

Option C is the correct option.

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Use the distributive property to remove the parentheses. <br> -5(-6w+3v-5)
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Answer:

30w - 15v + 25

Step-by-step explanation:

Multiply -5 by each term inside the parentheses.

-5(-6w+3v-5) =

= -5 \times (-6w) + (-5) \times 3v + (-5) \times (-5)

= 30w - 15v + 25

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The length of the longest side of a triangle must be shorter than the sum of the other two sides of the triangle. true or false
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This statement is true.

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21. Who is closer to Cameron? Explain.
pickupchik [31]

Problem 21

<h3>Answer:  Jamie is closer</h3>

-----------------------

Explanation:

  • A = Arthur's location = (20,35)
  • J = Jamie's location = (45,20)
  • C = Cameron's location = (65,40)

To find out who's closer to Cameron, we need to compute the segment lengths AC and JC. Then we pick the smaller of the two lengths.

We use the distance formula to find each length

Let's find the length of AC.

A = (x_1,y_1) = (20,35)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from A to C} = \text{length of segment AC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-65)^2 + (35-40)^2}\\\\d = \sqrt{(-45)^2 + (-5)^2}\\\\d = \sqrt{2025 + 25}\\\\d = \sqrt{2050}\\\\d \approx 45.2769257\\\\

The distance from Arthur to Cameron is roughly 45.2769257 units.

Let's repeat this process to find the length of segment JC

J = (x_1,y_1) = (45,20)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from J to C} = \text{length of segment JC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(45-65)^2 + (20-40)^2}\\\\d = \sqrt{(-20)^2 + (-20)^2}\\\\d = \sqrt{400 + 400}\\\\d = \sqrt{800}\\\\d \approx 28.2842712\\\\

Going from Jamie to Cameron is roughly 28.2842712 units

We see that segment JC is shorter than AC. Therefore, Jamie is closer to Cameron.

=================================================

Problem 22

<h3>Answer:  Arthur is closest to the ball</h3>

-----------------------

Explanation:

We have these key locations:

  • A = Arthur's location = (20,35)
  • J = Jamie's location = (45,20)
  • C = Cameron's location = (65,40)
  • B = location of the ball = (35,60)

We'll do the same thing as we did in the previous problem. This time we need to compute the following lengths:

  • AB
  • JB
  • CB

These segments represent the distances from a given player to the ball. Like before, the goal is to pick the smallest of these segments to find out who is the closest to the ball.

The steps are lengthy and more or less the same compared to the previous problem (just with different numbers of course). I'll show the steps on how to get the length of segment AB. I'll skip the other set of steps because there's only so much room allowed.

A = (x_1,y_1) = (20,35)\\\\B = (x_2,y_2) = (35,60)\\\\d = \text{Distance from A to B} = \text{length of segment AB}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-35)^2 + (35-60)^2}\\\\d = \sqrt{(-15)^2 + (-25)^2}\\\\d = \sqrt{225 + 625}\\\\d = \sqrt{850}\\\\d \approx 29.1547595\\\\

Segment AB is roughly 29.1547595 units.

If you repeated these steps, then you should get these other two approximate segment lengths:

JB = 41.2310563

CB = 36.0555128

-------------

So in summary, we have these approximate segment lengths

  • AB = 29.1547595
  • JB = 41.2310563
  • CB = 36.0555128

Segment AB is the smallest of the trio, which therefore means Arthur is closest to the ball.

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