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Kamila [148]
3 years ago
10

I REALLY NEED HELP WITH THIS PROBLEM! PICTURE ATTACHED

Mathematics
2 answers:
Anettt [7]3 years ago
7 0
C is the answer I just know
alexdok [17]3 years ago
7 0
It is C. 0.046x + 187.50
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How do you find the domain and range?
dedylja [7]

The Domain is always the x values and the range is alway the y values. For example (3,0) the domain is 3 and the range is 0. Hope this helps!

6 0
3 years ago
Carly and rob combined their dvd collection now they have 42 dvds all together. Carly had 4 more dvds than rob how many dvds did
Wewaii [24]
If Carly had more DVDs than Rob then you would subtract her amount away from the total. 42-4=38
4 0
3 years ago
How many distinguishable 7 letter​ "words" can be formed using the letters in ALABAMA​?
aksik [14]

Answer:ALABAMA has seven letters

Step-by-step explanation:

so we will start at 7! Counting the number of A's in the word we have 4 A's and so we will divide it by 4!

8 0
4 years ago
If you make $12 an hour and get a 10%decrease in pay what is your new hourly pay​
kondaur [170]

Answer:

$10.80

Step-by-step explanation:

1. Multiply $12 X 10%

2. convert 10% to decimal = .10

3. 12 X .10 = 1.20

4. Subtract 1.20 from 12

5. 12-1.20 = 10.80

6 0
3 years ago
Q4<br> Help pleaseeeeeeee
emmasim [6.3K]

a. The general equation for a circle centered at (a,b) with radius r is

(x-a)^2+(y-b)^2=r^2

The described circle has equation

(x+3)^2+(y+2)^2=r^2

We know the circle passes through the origin. This means that the equation above holds for x=0 and y=0. The distance between any point on the circle and its center is the radius, so we can use this fact to determine r:

(0+3)^2+(0+2)^2=r^2\implies 9+4=13=r^2\implies r=\sqrt{13}

So the circle's equation is

(x+3)^2+(y+2)^2=(\sqrt{13})^2=13

b. If the distance between point B and the center is less than \sqrt{13}, then B lies inside the circle. If the distance is greater than \sqrt{13}, it falls outside the circle. Otherwise, if the distance is exactly \sqrt{13}, then B lies on the circle.

The distance from B to the center is

\sqrt{(-1+3)^2+(3+2)^2}=\sqrt{4+25}=\sqrt{29}

29>13, so \sqrt{29}>\sqrt{13}, which means B falls outside the circle.

5 0
3 years ago
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