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Pavel [41]
3 years ago
9

To find the standard equation for a circle centered at the origin, we use the distance formula, since the radius measures?

Mathematics
2 answers:
Mrac [35]3 years ago
3 0
The distance from any point on the circle to the origin
____ [38]3 years ago
3 0

Answer: The answer is (a). the distance from any point in the circle to the origin.

Step-by-step explanation:  We know that to find the standard equation of a circle centred at the origin, we always use the distance formula. We are to select the correct measurement for the radius from the options given.

In a circle, as shown in the attached figure, the radius 'r' is the distance from any point 'P' on the circle to the centre 'O' of the circle. The centre of the circle in standard form is usually the origin.

So, we can say that radius measures the distance from any point on the circle to the origin.

Thus, (a) is the correct option.

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Find the sum of the arithmetic sequence. -1, 2, 5, 8, 11, 14, 17
muminat
56 is you answer you need.
8 0
3 years ago
What is 2x+3y=30 &amp; x+y=14<br> In elimination
puteri [66]

Answer:

x= 12 and y= 2

Step-by-step explanation:

First you would line up the equations so the x's and y's are on top of each other. Then you would multiply x+y=14 by 3 to give you 3x+3y=42. After that you subtract 2x-3y=30 and 3x+3y=42 to give you an answer of x=12. After that, you plug in x with 12 in the equation x+y=14. You subtract 12 from both sides to get an answer of 2. So ur solution is (12,2)

   

4 0
3 years ago
1. y = x2 + 8x + 15<br> Find the zeros of the function by rewriting the function in intercept form
lubasha [3.4K]

The zeros of given function y=x^{2}+8 x+15 is – 5 and – 3

<u>Solution:</u>

\text { Given, equation is } y=x^{2}+8 x+15

We have to find the zeros of the function by rewriting the function in intercept form.

By using intercept form, we can put value of y as  to obtain zeros of function

We know that, intercept form of above equation is x^{2}+8 x+15=0

\text { Splitting } 8 x \text { as }(5+3) x \text { and } 15 \text { as } 5 \times 3

\begin{array}{l}{\rightarrow x^{2}+(5+3) x+5 \times 3=0} \\\\ {\rightarrow x^{2}+5 x+3 x+5 \times 3=0}\end{array}

Taking “x” as common from first two terms and “3” as common from last two terms

x (x + 5) + 3(x + 5) = 0

(x + 5)(x + 3) = 0

Equating to 0 we get,

x + 5 = 0 or x + 3 = 0

x = - 5 or – 3

Hence, the zeroes of the given function are – 5 and – 3

5 0
3 years ago
Determine the value for b assuming that the two horizontal lines displayed are parallel.
Neko [114]

Answer:

29

Step-by-step explanation:

8 0
3 years ago
Help plZZZ ASAP thx
Kay [80]
512=2w^2 is the best answer
6 0
3 years ago
Read 2 more answers
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