Step-by-step explanation:
(3x+3)(x+2)=
3x^2+6x+3x+6
=3x^2+9x+6
we get quadratic equation
find discriminant:
D=b^2-4ac=81-72=9
x=(-b±√D)/2a
x1=-6/6=-1
x2=-12/6=-2
one of the x values should be correct.
but why when I try to put values I get zero
Answer: 1(a) $120
(ii a) $34
Step-by-step explanation:
21-19 = 2
$6/2=3
21x 3( one ratio =$3) =63
19x3 = 57
63+57=120 - (1 A ans)
(ii a) 40x0.15 (percentage decrease)= 6
40-6 = $34
<u>the correct question is</u>
The denarius was a unit of currency in ancient rome. Suppose it costs the roman government 10 denarii per day to support 4 legionaries and 4 archers. It only costs 5 denarii per day to support 2 legionaries and 2 archers. Use a system of linear equations in two variables. Can we solve for a unique cost for each soldier?
Let
x-------> the cost to support a legionary per day
y-------> the cost to support an archer per day
we know that
4x+4y=10 ---------> equation 1
2x+2y=5 ---------> equation 2
If you multiply equation 1 by 2
2*(2x+2y)=2*5-----------> 4x+4y=10
so
equation 1 and equation 2 are the same
The system has infinite solutions-------> Is a consistent dependent system
therefore
<u>the answer is</u>
We cannot solve for a unique cost for each soldier, because there are infinite solutions.
Excercise 1:
No, this is not simplified fully. The full answer would be <span>−7dk+14d−21k</span>−7, not <span>14d - 9 - 21k - 7dk + 2. So, it is not equivalent.
</span><span>
Excercise 4:
</span><span>7dk (2 - 3 - 1) - 7
</span>
Just multiply 7dk into (2) (-3) and (-1)
you'd get:
<span><span>−<span>14dk</span></span>+</span>−<span>7 after simplifying it fully.
</span>
p = parameter
w = width
w+2.75
p = 2w+2(w+2.75) = <span>4w</span>+<span>5.5 = 30
Answer to the last one:
w = 6.125</span>
