Dear Ffw299, we can find the circumference of a dime using the formula C=2πr. 2x3.14x8.5=59.5 millimeters in circumference.
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=\frac{x_1+x_2}{2} , b= \frac{y_1+y_2}{2})](https://tex.z-dn.net/?f=%28a%3D%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D%20%2C%20b%3D%20%5Cfrac%7By_1%2By_2%7D%7B2%7D%29)
a= 2.5, b= 3.5, x1= -2 and y1= 4
Plug in all the values and findout x2, y2
![(2.5=\frac{-2+x_2}{2} , 3.5= \frac{4+y_2}{2})](https://tex.z-dn.net/?f=%282.5%3D%5Cfrac%7B-2%2Bx_2%7D%7B2%7D%20%2C%203.5%3D%20%5Cfrac%7B4%2By_2%7D%7B2%7D%29)
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)
Step-by-step explanation:
can you please add details to your question
<span>mostly collect like terms
use associative property which is
(a+b)+c=a+(b+c)
also -a+b-c=(-a)+(b)+(-c) so you can move them around
and remember that:
you just use a general rule
x+x=2x
x^2+x^2=2x^2
3xy4xy=7xy
3x+4x^2=3x+4x^2
you
can only add like terms( like terms are terms that are same name like x
or y are differnt, and like terms have same power exg x^2 and x^3 and
x^1/2 and such
I will oly put the naswers because I don't have much time
first one: 2a+3b+2c
second one: remember that -(-6c)=+6c so the answer is c-10a-2b
third one: -a-8b-5c
</span>
Answer:
length = 5
width = 9
Step-by-step explanation:
V = lwh
let 'x' = length
let '4+x' = width
9 = height
405 = 9x(4+x)
405 = 36x + 9x²
this is a quadratic equation: 9x² + 36x - 405 = 0
factor out a 9 to get:
9(x² + 4x - 45) = 0
9(x-5)(x+9) = 0
one root is 5 and the other is -9
since distance cannot be negative we use 5 as the length and 9 is the width