104-32=72 brownies
72/32= 2.25
Ratio: 1 cookie:2.25 brownies
A :-) 1.) Given - base = 9 cm
height ( alt ) = 12 cm
hypotenuse ( hypo ) = x
Solution -
By Pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( x )^2 = ( 9 )^2 + ( 12 ) ^2
( x )^2 = 81 + 144
( x )^2 = 225
( x ) = _/225
( x ) = 15 cm
.:. The value of x ( hypotenuse ) = 15 cm
2.) Given - base = 10 cm
Height = 24 cm
Hypotenuse = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( x )^2 = ( 10 )^2 + ( 24 )^2
( x )^2 = 100 + 576
( x )^2 = 676
( x ) = _/676
( x ) = 26
.:. The value of x ( hypotenuse ) = 26 cm
3.) Given - base = 3 cm
Height = 7 cm
Hypotenuse = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( x )^2 = ( 3 )^2 + ( 7 )^2
( x )^2 = 9 + 49
( x )^2 = 58
( x ) = _/58
( x ) = 7.6
.:. The value of x ( hypotenuse ) = 7.6 cm
4.) Given - base = 10 cm
Height = 6 cm
Hypotenuse = x
Solution -
By pythagorus theorem
( Hypo )^2 = ( base )^2 + ( alt )^2
( x )^2 = ( 10 )^2 + ( 6 )^2
( x )^2 = 100 + 36
( x )^2 = 136
( x ) = _/136
( x ) = 11.6
.:. The value of x ( hypotenuse ) = 11.6 cm
5.) Given - hypotenuse = 24 cm
height = 6 cm
Base = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( 24 )^2 = ( x )^2 + ( 6 )^2
( x )^2 = ( 6 )^2 - ( 24 )^2
( x )^2 = 36 - 576
( x )^2 = -540
( x ) = _/-540
( x ) = 23.2
.:. The value of x ( base ) = 23.2 cm
6.) Given - base = 1 cm
height = 1 cm
hypotenuse = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( x )^2 = ( 1 )^2 + ( 1 )^2
( x )^2 = 1 + 1
( x )^2 = 2
( x ) = _/2
( x ) = 1.4
.:. The value of x ( hypotenuse ) = 1.4 cm
7.) Given - hypotenuse = 21 cm
height = 8 cm
Base = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( 21 )^2 = ( x )^2 + ( 8 )^2
441 = ( x )^2 + 64
( x )^2 = 64 - 441
( x )^2 = -377
( x ) = _/-377
( x ) = 19.4
.:. The value of x ( base ) = 19.4
8.) given - height = 24 cm
Hypotenuse = 30cm
Base = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( 30 )^2 = ( x )^2 + ( 24 )^2
900 = ( x )^2 + 576
( x )^2 = 576 - 900
( x )^2 = -324
( x ) = _/-324
( x ) = 18
.:. The value of x ( base ) = 18 cm
9.) ( i ) lets find ‘x’
Given - base = 9 cm
height = 5 cm
hypotenuse = x
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( x )^2 = ( 9 )^2 + ( 5 )^2
( x )^2 = 81 +25
( x )^2 = 106
( x ) = _/106
( x ) = 10.2
.:. The value of x ( hypotenuse )
= 10.2 cm
( ii ) lets find ‘y’
Given - base = 3 cm
height = 5 cm
Hypotenuse = y
Solution -
By pythagorus theorem
( hypo )^2 = ( base )^2 + ( alt )^2
( y )^2 = ( 3 )^2 + ( 5 )^2
( y )^2 = 9 + 25
( y )^2 = 34
( y ) = _/34
( y ) = 5.8
.:. The value of y ( hypotenuse )
= 5.8 cm
Ok so assumin by 2x3 you mean 2x^3 or 2 times x cubed
PEMDAS
parenthasees
exponentes
multiply
divide
add
subtract
exponents
we have
2(3)^3+3(4)^2-17
exonents
3^3=27
4^2=16
now we have
2(27)+3(16)-17
distribute (multply)
2 tiems 27=54
3 times 16=48
now we have
54+48-17
add/subtract
85
answer is 85
<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.
