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densk [106]
2 years ago
12

What are the center and radius of the circle with equation (x + 8)^2 + (y + 3)^2 = 121?

Mathematics
1 answer:
I am Lyosha [343]2 years ago
7 0
6 hould be the answer
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Daria bought a bracelet at original cost $16 to sell in her handicraft store. She marked the price up 45%.
KengaRu [80]
$23.20. 45% of $16 is 7.2
6 0
3 years ago
5= 3 1 (x+7) ) x = <br> What is it?
aleksandrvk [35]

Answer:

Step-by-step explanation:

5= 31(x +7)

5= 31x + 217

5 - 217 = 31x + 217 - 217

212 = 31x

212/31 = 31x/31

212 = x

x= 212

5 0
2 years ago
A cone with a radius of 6 cm has a volume of 1200 cm3. What is the height of the cone? (The volume of a cone is given by the for
Rina8888 [55]

Answer:

  • 31.8 cm

Step-by-step explanation:

<u>Volume of a cone:</u>

  • V = 1/3πr²h

<u>We have:</u>

  • V = 1200 cm³
  • r = 6 cm
  • h = ?

<u>Substitute values and solve for h:</u>

  • 1200 = 1/3*3.14*6²h
  • 1200 = 37.68h
  • h = 1200/37.68
  • h = 31.8 cm (rounded)

None of the choices is correct

5 0
3 years ago
Read 2 more answers
Looking at the two quadratic functions below (1 &amp; 2), answer the following questions.
algol [13]
Part A:

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that parabola (2) is stretched horizontally by a factor of 13 which is greater than 1. This means that parabola (2) is further away from the x-axis than parabola (1). (i.e. parabola (2) is more 'vertical' than parabola (1).

Therefore, parabola (1) is wider than parabola (2).



Part B:

A parabola open up when the coefficient of the quadratic term (the squared term) is positive and opens down when the coefficient of the quadratic term is negative.

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that the coefficient of the quadratic term is positive for parabola (2) and negative for parabola (1), therefore, the parabola that will open down is parabola (1).



Part C:

For any function, f(x), the graph of the function is moved p places to the left when p is added to x (i.e. f(x + p)) and moves p places to the right when p is subtracted from x (i.e. f(x - p)).

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that in parabola (1), 12 is added to x, which means that the graph of the parent function is shifted 12 places to the left while in parabola (2), 4 is subtracted from x, which means that the graph of the parent function is shifted 4 places to the right.

Therefore, the parabola that would be furthest left on the x-axis is parabola (1).



Part D:

For any function, f(x), the graph of the function is moved q places up when q is added to the function (i.e. f(x) + q) and moves q places down when q is subtracted from the function (i.e. f(x) - q).

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that in parabola (2), 1 is added to the function, which means that the graph of the parent function is shifted 1 place up while in parabola (1), 6 is subtracted from the function, which means that the graph of the parent function is shifted 6 places down.

Therefore, the parabola that would be highest on the y-axis is parabola (2).
7 0
3 years ago
I need help with my math homework it is algebra
Ivahew [28]
What are the math problems?
6 0
2 years ago
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