Which parallelogram will have the smallest area?
2 answers:
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Point a is at (-2,4) => (x1,y1)
Midpoint is (2.5, 3.5) => (a,b)
Point B is (x2, y2)
To find midpoint we use formula
a= 2.5, b= 3.5, x1= -2 and y1= 4
Plug in all the values and findout x2, y2
multiply 2 on both sides to remove fraction
(5 = -2+x2 , 7 = 4+ y2)
5 = -2+x2, so x2= 7
7 = 4+ y2, so y2= 3
The point B is ( 7, 3)
Easy, it says "for ANY positive integer" so just test any positive integer
remember that n! means times all integers from 1 to that number n
lets try 1
(1+1)!/(1!)-1=
(2!)/(1!)-1=
2/1-1=
2-1=
1
if you don't believe me, try 2
(2+1)!/(2!)-2=
(3!)/(2!)-2=
(6)/(2)-2=
3-2=
1
te answer is 1, B
and number 9
easy, remember the exponential law
(x^m)(x^n)=x^(m+n)
jsut add the exponents
first gropu like bases
(r^2r^2/3)(t^1/2t^-3/2)
add bases
(r^2 and 2/3)(t^-1)=
(r^2∛(r^2))(1/t)=
![\frac{r^{2} \sqrt[3]{r^{2}} }{t}](https://tex.z-dn.net/?f=%20%5Cfrac%7Br%5E%7B2%7D%20%5Csqrt%5B3%5D%7Br%5E%7B2%7D%7D%20%7D%7Bt%7D%20)
remember that n! means times all integers from 1 to that number n
lets try 1
(1+1)!/(1!)-1=
(2!)/(1!)-1=
2/1-1=
2-1=
1
if you don't believe me, try 2
(2+1)!/(2!)-2=
(3!)/(2!)-2=
(6)/(2)-2=
3-2=
1
te answer is 1, B
and number 9
easy, remember the exponential law
(x^m)(x^n)=x^(m+n)
jsut add the exponents
first gropu like bases
(r^2r^2/3)(t^1/2t^-3/2)
add bases
(r^2 and 2/3)(t^-1)=
(r^2∛(r^2))(1/t)=