0.3(x - 2y) + 0.5x - y
0.3x - 0.6y + 0.5x - y
(0.3x + 0.5x) + (-0.6y - y)
0.8x + 1.6y
This means the answer is B.
Feel free to ask me any questions about method of solving in the comments :)
Answer: i don,t know sorry but i wish you the best of luck
Step-by-step explanation:
Answer:
(2.5 , 3.5)
Step-by-step explanation:
We can use the midpoint formula . Here the points are , (2,2) and (3,5) .
• <u>Using</u><u> </u><u>Midpo</u><u>int</u><u> Formula</u><u> </u><u>:</u><u>-</u><u> </u>
⇒ M = { (x1 + x2)/2 , (y1 + y2)/2 }
⇒ M = ( 2+3/2 , 5+2/2 )
⇒ M = ( 5/2 , 7/2 )
⇒ M = ( 2.5 , 3.5 )
<h3>
<u>Hence </u><u>the</u><u> </u><u>midpoint</u><u> </u><u>is</u><u> </u><u>(</u><u>2</u><u>.</u><u>5</u><u> </u><u>,</u><u> </u><u>3</u><u>.</u><u>5</u><u>)</u></h3>
Find numbers that multiply to 28 and add them to see if they add to 8
28=
1 and 28=29 not 8
2 and 14=16 not 8
4 and 7=11 not 8
that's it'
no 2 numbers
we must use quadratic formula
x+y=8
xy=28
x+y=8
subtract x fromb oths ides
y=8-x
subsitute
x(8-x)=28
distribute
8x-x^2=28
add x^2 to both sides
8x=28+x^2
subtract 8x
x^2-8x+28=0
if you have
ax^2+bx+c=0 then x=

so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28
x=

x=

x=

x=

x=

there are no real numbers that satisfy this