Answer:
Option C
Step-by-step explanation:
Standard equation:
r^2 = (x-h)^2 + (y-k)^2
Radius: r
Center: (h,k)
Plug the values in.
r^2 = (x+8)^2 + (y-15)^2
Using distance formula, you conclude that the radius is 17 since the distance from (-8,15) to (0,0) is 17.
r^2 = 289
That means option C is the right answer.
289 = (x+8)^2 + (y-15)^2
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Your answer is A.
why:
because if the y value changes then your vertical will change. if your x changes than your horizontal will change. if it’s compressed, your x value is less than 1.
Area = (base)*(height)
15 = (x+7)(x-7)
(x+7)(x-7) = 15
x^2 - 49 = 15
x^2 - 49+49 = 15+49
x^2 = 64
sqrt(x^2) = sqrt(64)
x = 8 ... note that x can't be negative
If x = 8, then the base is...
x+7 = 8+7 = 15
Therefore the base is 15 units.
(the height is x-7 = 8-7 = 1 unit)
Answer:
27°
Step-by-step explanation:
Since BD is a diameter then arc BD = 180°
∠ APE = 180° - (90 + 63)° = 180° - 153° = 27°
The measure of the arc AE is equal to the angle at the centre subtended by it.
arc AE = ∠ APE = 27°
The data below shows the average number of text messages sent daily by a group of people: 7, 8, 4, 7, 5, 2, 5, 4, 5, 7, 4, 8, 2,
enot [183]
It all depends. You've given us an incredibly vague question.
The outlier could be a number that's low or quite high. Also, outliers
shouldn't really contribute towards the value of the mean, median or
range related to a group of data.
They are called outliers because they are bizarre results or numbers
and should be detached from groups of data. Outliers by definition
are abnormalities or anomalies.
I'd say outliers don't really change anything, unless you actually want
to give them credibility or weight.
Large outliers can inflate the value of means, medians and ranges.
Small outliers will invariably deflate the value of means and medians.
