Hello!
As you can see this line is horizontal. It is 6 units long. Therefore the center is three units in. Therefore the center of this circle is at (1,2).
I hope this helps!
<h3>given:</h3>

<h3>to find:</h3>
the radius of the given ball (sphere).
<h3>solution:</h3>
![r = \sqrt[3]{ \frac{3v}{4\pi} }](https://tex.z-dn.net/?f=r%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3v%7D%7B4%5Cpi%7D%20%7D%20)
![r = \sqrt[3]{ \frac{3 \times 905}{4\pi} }](https://tex.z-dn.net/?f=r%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3%20%5Ctimes%20905%7D%7B4%5Cpi%7D%20%7D%20)

<u>therefore</u><u>,</u><u> </u><u>the</u><u> </u><u>radius</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>ball</u><u> </u><u>is</u><u> </u><u>6</u><u> </u><u>cm</u><u>.</u>
note: refer to the picture I added on how you can change r as the subject of the formula.
Step-by-step explanation:
You can solve systems of equations using either substitution or elimination. For these problems, I recommend elimination. I'll do the first one as an example.
-3x + 16y = 9
-4x + 8y = 12
Multiply the second equation by -2.
8x − 16y = -24
Add to the first equation (notice the y's cancel out).
(-3x + 16y) + (8x − 16y) = 9 − 24
5x = -15
Solve for x.
x = -3
Now you can plug this into either equation to find y.
-3(-3) + 16y = 9
9 + 16y = 9
y = 0
The solution is (-3, 0).
Answer:
A. (x-5)(x-7)
B. It means the zeroes of the graph are at (5,0) and (7,0)
Step-by-step explanation:
A
find what + what equals -12, and multiples to 35.
List all multiples of 35.
1 35
<em>5 7</em>
- 12 is negative so that means it is -5 and -7
(x-5)(x-7)
B
The factor numbers are where the zeroes on a graph are, <u>but</u> whatever is in parenthesis, the number is opposite on a graph (e.g. if it was (x-3), the zero would be (3,0)
We can not really tell in this question as you dont know the equation that is being used for the domain and range relationship but overall one should know that:
The set of values of the independent variable(s) for which a function or relation is defined as the domain of a function. Typically, this is the set of x-values that give rise to real y-values.
The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain.