Answer:
yes
Step-by-step explanation:
5 times 3 is 15 and 12 and 15 and 12 divided by 3 you will get 5/4
Hello,
Let's place the last digit: it must be 2 or 4 or 8 (3 possibilities)
It remainds 4 digits and the number of permutations fo 4 numbers is 4!=4*3*2*1=24
Thus there are 3*24=72 possibilities.
Answer A
If you do'nt believe run this programm
DIM n(5) AS INTEGER, i1 AS INTEGER, i2 AS INTEGER, i3 AS INTEGER, i4 AS INTEGER, i5 AS INTEGER, nb AS LONG, tot AS LONG
tot = 0
n(1) = 1
n(2) = 2
n(3) = 4
n(4) = 7
n(5) = 8
FOR i1 = 1 TO 5
FOR i2 = 1 TO 5
IF i2 <> i1 THEN
FOR i3 = 1 TO 5
IF i3 <> i2 AND i3 <> i1 THEN
FOR i4 = 1 TO 5
IF i4 <> i3 AND i4 <> i2 AND i4 <> i1 THEN
FOR i5 = 1 TO 5
IF i5 <> i4 AND i5 <> i3 AND i5 <> i2 AND i5 <> i1 THEN
nb = ((((n(i1) * 10) + n(i2)) * 10 + n(i3)) * 10 + n(i4)) * 10 + n(i5)
IF nb MOD 2 = 0 THEN
tot = tot + 1
END IF
END IF
NEXT i5
END IF
NEXT i4
END IF
NEXT i3
END IF
NEXT i2
NEXT i1
PRINT "tot="; tot
END
Answer:
<em>Answer is</em><em> </em><em>1345</em><em>:</em><em>12</em>
Step-by-step explanation:

<em>HAVE A NICE DAY</em><em>!</em>
<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>
Let
A------> <span>(5√2,2√3)
B------> </span><span>(√2,2√3)
we know that
</span>the abscissa<span> and the ordinate are respectively the first and second coordinate of a point in a coordinate system</span>
the abscissa is the coordinate x<span>
step 1
find the midpoint
ABx------> midpoint AB in the coordinate x
</span>ABy------> midpoint AB in the coordinate y
<span>
ABx=[5</span>√2+√2]/2------> 6√2/2-----> 3√2
ABy=[2√3+2√3]/2------> 4√3/2-----> 2√3
the midpoint AB is (3√2,2√3)
the answer isthe abscissa of the midpoint of the line segment is 3√2
see the attached figure
It’s an isosceles triangle, so those two sides are equal:
3x-15=x+33
2x=48
x=24, so BC should be 24.