Hello :
an equation is : (x+2)² +(y-7)² = 4
x² +4x+4 +y² -14y +49 = 4
x² +4x+y² -14y +49 = 0 ....( answer : 3)
10=-x-8-6
10=-x-14
Add x to each side
10+x=-x-14+x
10+x=-14
Subtract 10 to each side
10+x-10=-14-10
x=-24
Check:
10=-x-8-6
Substitute x with -24
10=-(-24)-8-6
10=24-8-6
10=16-6
10=10. As a result, x=-24. Hope it help!
2f(x) = 2x - 4 [0, 3]3f(x) = 3x - 1 [-2, -1]9f(x) = x² [4, 5]4f(x) = 4x [5, 20]5f(x) = x² - 3 [0, 5]1f(x) = x + 10 [-5, -1]10f(x) = 10x [-3, 0]¹/₂f(x) = 0.5x - 2 [2, 4]11f(x) = 2x² + x [1, 4]-1f(x) = -x + 2 [-3, 5]Domainthe set of all reasonable input values of x for the functionRangeset of output y values for the domain of the functionAverage Rate of ChangeChange in values over a given interval.Origin(0,0) on the coordinate graphing system; where the two axes meetx-axisthe horizontal number line in the coordinate systemy-axisthe vertical number line in the coordinate systemCoordinatesany specific (x,y) in the coordinate systemx-interceptwhere the function intersects the x-axisy-interceptwhere the function intersects the y-axis; the b value in a linear functionLinear FunctionA function whose graph is a straight line, where the average rate of change (slope) is constant.Exponential FunctionA function where the average rate of change is not constant and whose input value is an exponent.Table of ValuesA table showing two sets of related numbers<span>Slope of line through the points (-2, 3) and (0,0)
m = (0 - 3) / (0 - -2) = -3/2</span><span>Average Rate of Change on the interval
[-2, 0]</span>Slope: m = "rise over run" = 2Rate of Change<span>Slope of line through the points (5, -1) and (0,0)
m = (0 - -1) / (0 - 5) = -1/5</span><span>Average Rate of Change on the interval
[0, 5]</span><span>Slope of line through the points
(0, 16) and (4, 21)
m = (21 - 16) / (4 - 0) = 5/4</span>Average Rate of Change over the interval [0,4]
The summand (R?) is missing, but we can always come up with another one.
Divide the interval [0, 1] into 
subintervals of equal length 
:
![[0,1]=\left[0,\dfrac1n\right]\cup\left[\dfrac1n,\dfrac2n\right]\cup\cdots\cup\left[1-\dfrac1n,1\right]](https://tex.z-dn.net/?f=%5B0%2C1%5D%3D%5Cleft%5B0%2C%5Cdfrac1n%5Cright%5D%5Ccup%5Cleft%5B%5Cdfrac1n%2C%5Cdfrac2n%5Cright%5D%5Ccup%5Ccdots%5Ccup%5Cleft%5B1-%5Cdfrac1n%2C1%5Cright%5D)
Let's consider a left-endpoint sum, so that we take values of 
where 
is given by the sequence
with 
. Then the definite integral is equal to the Riemann sum
Answer:
same
Step-by-step explanation:
