Answer:
162.4 in²
Step-by-step explanation:
LETS GET INTOOOOEEETTT
Let's start with what we know:
Area of regular octagon = 1/2 x perimeter x apothem
We know the apothem, so all that we need to find to fill in the above equation is the perimeter:
perimeter = 8 x 5.8 = 46.4in
Now we can fill in our original equation and solve:
Area of regular octagon = 1/2 x perimeter x apothem
Formula = n (s/2)² divided by tan( π /n)
= 8 (5.8/2)² divided by tan ( π /8)
= 162.4283 in²
ORRR when rounded to the nearest tenth,
=162.4 in²
Height=4 ft
angle=33 degrees
My tip first is to draw it out. Draw an acute angle opening up to a building forming a triangle. Inside of the vertex of the acute angle is the degree 33 and the height of the building is 4 ft. So, use trig! Do you have a calculator? Opposite over hypotenuse is Sine. (SOH CAH TOA). (4)/(sin(33)). which equals 7.34 ft as your height! Hope this helped!
Let <em>q</em> be the number of quarts of pure antifreeze that needs to be added to get the desired solution.
8 quarts of 40% solution contains 0.40 × 8 = 3.2 quarts of antifreeze.
The new solution would have a total volume of 8 + <em>q</em> quarts, and it would contain a total amount of 3.2 + <em>q</em> quarts of antifreeze. You want to end up with a concentration of 60% antifreeze, which means
(3.2 + <em>q</em>) / (8 + <em>q</em>) = 0.60
Solve for <em>q</em> :
3.2 + <em>q</em> = 0.60 (8 + <em>q</em>)
3.2 + <em>q</em> = 4.8 + 0.6<em>q</em>
0.4<em>q</em> = 1.6
<em>q</em> = 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
