Answer:
<u>multiply</u> the base and <u>add </u>the exponents
Answer:
![\large\boxed{y=\dfrac{1}{4}x^2-x-4}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7By%3D%5Cdfrac%7B1%7D%7B4%7Dx%5E2-x-4%7D)
Step-by-step explanation:
The equation of a parabola in vertex form:
![y=a(x-h)^2+k](https://tex.z-dn.net/?f=y%3Da%28x-h%29%5E2%2Bk)
<em>(h, k)</em><em> - vertex</em>
The focus is
![\left(h,\ k+\dfrac{1}{4a}\right)](https://tex.z-dn.net/?f=%5Cleft%28h%2C%5C%20k%2B%5Cdfrac%7B1%7D%7B4a%7D%5Cright%29)
We have the vertex (2, -5) and the focus (2, -4).
Calculate the value of <em>a</em> using ![k+\dfrac{1}{4a}](https://tex.z-dn.net/?f=k%2B%5Cdfrac%7B1%7D%7B4a%7D)
<em>k = -5</em>
<em>add 5 to both sides</em>
<em>multiply both sides by 4</em>
![4\!\!\!\!\diagup^1\cdot\dfrac{1}{4\!\!\!\!\diagup_1a}=4](https://tex.z-dn.net/?f=4%5C%21%5C%21%5C%21%5C%21%5Cdiagup%5E1%5Ccdot%5Cdfrac%7B1%7D%7B4%5C%21%5C%21%5C%21%5C%21%5Cdiagup_1a%7D%3D4)
![\dfrac{1}{a}=4\to a=\dfrac{1}{4}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Ba%7D%3D4%5Cto%20a%3D%5Cdfrac%7B1%7D%7B4%7D)
Substitute
![a=\dfrac{1}{4},\ h=2,\ k=-5](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B1%7D%7B4%7D%2C%5C%20h%3D2%2C%5C%20k%3D-5)
to the vertex form of an equation of a parabola:
![y=\dfrac{1}{4}(x-2)^2-5](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B1%7D%7B4%7D%28x-2%29%5E2-5)
The standard form:
![y=ax^2+bx+c](https://tex.z-dn.net/?f=y%3Dax%5E2%2Bbx%2Bc)
Convert using
![(a-b)^2=a^2-2ab+b^2](https://tex.z-dn.net/?f=%28a-b%29%5E2%3Da%5E2-2ab%2Bb%5E2)
![y=\dfrac{1}{4}(x^2-2(x)(2)+2^2)-5\\\\y=\dfrac{1}{4}(x^2-4x+4)-5](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B1%7D%7B4%7D%28x%5E2-2%28x%29%282%29%2B2%5E2%29-5%5C%5C%5C%5Cy%3D%5Cdfrac%7B1%7D%7B4%7D%28x%5E2-4x%2B4%29-5)
<em>use the distributive property: a(b+c)=ab+ac</em>
![y=\left(\dfrac{1}{4}\right)(x^2)+\left(\dfrac{1}{4}\right)(-4x)+\left(\dfrac{1}{4}\right)(4)-5\\\\y=\dfrac{1}{4}x^2-x+1-5\\\\y=\dfrac{1}{4}x^2-x-4](https://tex.z-dn.net/?f=y%3D%5Cleft%28%5Cdfrac%7B1%7D%7B4%7D%5Cright%29%28x%5E2%29%2B%5Cleft%28%5Cdfrac%7B1%7D%7B4%7D%5Cright%29%28-4x%29%2B%5Cleft%28%5Cdfrac%7B1%7D%7B4%7D%5Cright%29%284%29-5%5C%5C%5C%5Cy%3D%5Cdfrac%7B1%7D%7B4%7Dx%5E2-x%2B1-5%5C%5C%5C%5Cy%3D%5Cdfrac%7B1%7D%7B4%7Dx%5E2-x-4)
Answer:
Rational.
Step-by-step explanation:
![\textbf{A rational number is a real number that can be expressed as}~ \dfrac{p}{q},\\\\\textbf{where}~ p~ \textbf{and}~ q~ \textbf{are integers.}\\\\ \boxed{50.2 = \dfrac{502}{10}}\\\\ \textbf{50.2 can be expressed as a ratio of two integers,}~ \textbf{so it is a rational number.}](https://tex.z-dn.net/?f=%5Ctextbf%7BA%20rational%20number%20is%20a%20real%20number%20that%20can%20be%20expressed%20as%7D~%20%5Cdfrac%7Bp%7D%7Bq%7D%2C%5C%5C%5C%5C%5Ctextbf%7Bwhere%7D~%20p~%20%5Ctextbf%7Band%7D~%20q~%20%5Ctextbf%7Bare%20integers.%7D%5C%5C%5C%5C%20%5Cboxed%7B50.2%20%3D%20%5Cdfrac%7B502%7D%7B10%7D%7D%5C%5C%5C%5C%20%5Ctextbf%7B50.2%20can%20be%20expressed%20as%20a%20ratio%20of%20two%20integers%2C%7D~%20%5Ctextbf%7Bso%20it%20is%20a%20rational%20number.%7D)
AB is bisected by CD (TRUE). This is True because E is the midpoint between A and B and CD passes through E
CD is bisected by AB (FALSE) CD is bisected by point F and not AB
AE = 1/2 * AB (TRUE) since E is the midpoint of AB , E divides AB into two equal halves
EF = 1/2 * ED (FALSE) The true statement would have been CF = 1/2* CD
FD = EB (FALSE) sinc we do not know if CD and AB are of the same lengths
CE + EF = ED (TRUE) since F is the midpoint the sum of CE and EF is equal to ED
Answer:
The midpoint is ( 7,5)
Step-by-step explanation:
To find the x coordinate of the midpoint, add the x coordinate of the endpoints and divide by 2
( 4+10)/2 = 14/2 = 7
To find the y coordinate of the midpoint, add the y coordinate of the endpoints and divide by 2
(8+2)/2 = 10/2 = 5
The midpoint is ( 7,5)