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Margaret [11]
4 years ago
7

The endpoints of a segment are (0, –5) and (–3, –7). What are the endpoints of the segment after it has been translated 5 units

to the right?
A.(–5, –5), (–8, –7)
B.(0, 0), (–3, –2)
C.(5, –5), (2, –7)
D.(5, 0), (2, –2)
Mathematics
1 answer:
valkas [14]4 years ago
6 0
The answer is
<span>D.(5, 0), (2, –2)
proof
let A(0, -5) and </span>
<span>it has been translated 5 units to the right means, x=xA+5=0+5=5, y=yA+5=5-5=0</span>
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velikii [3]

Answer:

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Step-by-step explanation:

Name using degree: just count the highest number of variables in a single term. if it's 0, its a constant polynomial, if it's 1 it's linear, 2 quadratic, 3 cubic

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6 0
2 years ago
Find an explicit rule for the nth term of the arithmetic sequence.<br><br> -13, -7, -1, 5, ...
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-13 + 5*n-1 I think

Step-by-step explanation:

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8 0
3 years ago
Find the radius of the circle with equation x²+y²+8x+8y+28=0
galina1969 [7]
<h2>Hello!</h2>

The answer is:

Center: (-4,-4)

Radius: 2 units.

<h2>Why?</h2>

To solve the problem, using the given formula of a circle, we need to find its standard equation form which is equal to:

(x-h)^{2}+(y-k)^{2}=r^{2}

Where:

"h" and "k" are the coordinates of the center of the circle and "r" is its radius.

So, we need to complete the square for both variable "x" and "y".

The given equation is:

x^{2}+y^{2}+8x+8y+28=0

So, solving we have:

x^{2}+y^{2}+8x+8y=-28

(x^2+8x+(\frac{8}{2})^{2})+(y^2+8y+(\frac{8}{2})^{2})=-28+((\frac{8}{2})^{2})++(\frac{8}{2})^{2})\\\\(x^2+8x+16 )+(y^2+8y+16)=-28+16+16\\\\(x^2+4)+(y^2+4)=4

(x^2-(-4))+(y^2-(-4))=4

Now, we have that:

h=-4\\k=-4\\r=\sqrt{4}=2

So,

Center: (-4,-4)

Radius: 2 units.

Have a nice day!

Note: I have attached a picture for better understanding.

3 0
3 years ago
Help me please !!!!! It’s urgent
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I did that equation on the side and I got 442, I hope this helps
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