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lara31
17 days ago
6

The vet told Jake that his dog, Rocco, who weighed 55 pounds, needed to lose 10 pounds. Jake started walking Rocco every day and

changed the amount of food he was feeding him. Rocco lost half a pound the first week. Jake wants to determine Rocco’s weight in pounds, p, after w weeks if Rocco continues to lose weight based on his vet’s advice.
The equation of the scenario is what.

The values of p must be what.
Mathematics
2 answers:
tester [3.9K]17 days ago
6 0

Answer:

1. The equation that describes the situation is p = 55 - 0.5w

2. The possible values of p range from 45 to 55 as whole numbers

tester [3.9K]17 days ago
4 0

Answer:

0.5 times W

Step-by-step explanation:

Multiply 0.5 by the number of weeks W to find the total amount lost.

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In the diagram below BD is parallel to XY. What is the value of x
Svet_ta [4321]

The value of x equals 60 degrees.

This is because alternate interior angles are equal by definition :)

Can I get the brainliest award please?

5 0
14 days ago
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The pizza shop offers a 15 percent discount for veterans and senior citizens. If the price of a pizza is $12, how would you find
Svet_ta [4321]
To calculate, simply multiply 12 by 0.15 or 15%, which equals 1.8. Then subtract 1.8 from 12, yielding $10.20 as the discounted price for the pizza.

I hope this is helpful


8 0
11 days ago
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What other points are on the line of direct variation through (5, 12)? Check all that apply. (0, 0) (2.5, 6) (3, 10) (7.5, 18) (
Inessa [3907]

It is established that

A correlation between two variables, x and y, demonstrates a direct variation if it can be written in the format y/x=k or y=kx

For this question, we have

the point (5,12) lies on the direct variation line

therefore

Determine the constant of proportionality k

y/x=k-------> substitute ------> k=12/5

The equation is

y=\frac{12}{5}x

Keep in mind that

If a point is located on the direct variation line

then

the point has to fulfill the direct variation equation

we will now validate each point

case A) point (0,0)

x=0\ y=0

Insert the values of x and y into the direct variation equation

0=\frac{12}{5}*0

0=0 -------> is valid

thus

the point (0,0) lies on the direct variation line

case B) point (2.5,6)

x=2.5\ y=6

Insert the values of x and y into the direct variation equation

6=\frac{12}{5}*2.5

6=6 -------> is valid

thus

the point (2.5,6) lies on the direct variation line

case C) point (3,10)

x=3\ y=10

Insert the values of x and y into the direct variation equation

10=\frac{12}{5}*3

10=7.2 -------> is not valid

thus

the point (3,10) does not lie on the direct variation line

case D) point (7.5,18)

x=7.5\ y=18

Insert the values of x and y into the direct variation equation

18=\frac{12}{5}*7.5

18=18 -------> is valid

thus

the point (7.5,18) lies on the direct variation line

case E) point (12.5,24)

x=12.5\ y=24

Insert the values of x and y into the direct variation equation

24=\frac{12}{5}*12.5

18=30 -------> is not valid

thus

the point (12.5,24) does not lie on the direct variation line

case F) point (15,36)

x=15\ y=36

Insert the values of x and y into the direct variation equation

36=\frac{12}{5}*15

36=36 -------> is valid

thus

the point (15,36) lies on the direct variation line

ultimately

the solution is

(0,0)

(2.5,6)

(7.5,18)

(15,36)


5 0
1 day ago
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Lonzell said the function shown in the graph is positive on the interval (−,) and negative on the interval (−,−)open negative 5
Inessa [3907]

Answer:

The segments of the positive graph are located at the coordinates (-5,-4) and (2,5).

In contrast, the negative segment is identified as (-4,2).

Step-by-step explanation:

The positive graph is above the x-axis, while the negative portion lies beneath the y-axis.

Within the interval (-1,5),

The graph appears below the x-axis between (-1,2)

And above it between (2,5).

At this juncture, Lonzell’s assertion is inaccurate.

Within the interval (-5,-1),

The graph again resides below the x-axis in the range (-4,-1)

And above it between (-5,4).

At this point, Lonzell's assessment remains incorrect.

Thus,

The segments of the positive graph are at the coordinates (-5,-4) and (2,5).

Meanwhile, the negative segment is identified at (-4,2).

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7 days ago
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PIT_PIT [3919]
THE CIRCLE EQUATION: (x - h)² + (y - k)² = r²


= (x + 1)² + (y - 4)² = 3
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13 days ago
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