What is the perimeter of a square with a length of 19 meters?
2 answers:
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Hope this is the right one.
Problem 8
p^2 - 30p + c
<em>Step One</em>
Take 1/2 of - 30
1/2 * -30 = - 15
<em>Step 2
</em>Square -15
(-15)^2 = 225
c = 225
Problem Nine
a = 1
b = 4
c = -15
x = [-4 +/- sqrt(76)] / 2
x = [-4 +/- 2*sqrt19]/2
x = [-4/2 +/- 2/2 sqrt[19]
x = - 2 +/- sqrt(19)
x1 = - 2 + sqrt(19)
x2 = -2 - sqrt(19)
These two can be broken down more by finding the square root. I will leave them the way they are. It's just a calculator question if you want it to go into decimal form.
Problem Ten
a = 1
b = 4
c = -32
The discriminate is sqrt(b^2 - 4ac)
D = sqrt(b^2 - 4ac)
D = sqrt(4^2 - 4(1)(-32)
D = sqrt(16 - - 128)
D = sqrt(16 + 128)
D = sqrt(144)
D = +/- 12
Since D can equal + or minus 12 there must be 2 possible (and different) roots. As a matter of fact, this quadratic can be factored.
(x + 8)(x - 4) = y
But that' s not what you were asked for.
The discriminate is > 0 so the roots are going to be real.
<em>Answer; The discriminate is > 0 so there will be 2 real different roots.</em>
Problem 8
p^2 - 30p + c
<em>Step One</em>
Take 1/2 of - 30
1/2 * -30 = - 15
<em>Step 2
</em>Square -15
(-15)^2 = 225
c = 225
Problem Nine
a = 1
b = 4
c = -15
x = [-4 +/- sqrt(76)] / 2
x = [-4 +/- 2*sqrt19]/2
x = [-4/2 +/- 2/2 sqrt[19]
x = - 2 +/- sqrt(19)
x1 = - 2 + sqrt(19)
x2 = -2 - sqrt(19)
These two can be broken down more by finding the square root. I will leave them the way they are. It's just a calculator question if you want it to go into decimal form.
Problem Ten
a = 1
b = 4
c = -32
The discriminate is sqrt(b^2 - 4ac)
D = sqrt(b^2 - 4ac)
D = sqrt(4^2 - 4(1)(-32)
D = sqrt(16 - - 128)
D = sqrt(16 + 128)
D = sqrt(144)
D = +/- 12
Since D can equal + or minus 12 there must be 2 possible (and different) roots. As a matter of fact, this quadratic can be factored.
(x + 8)(x - 4) = y
But that' s not what you were asked for.
The discriminate is > 0 so the roots are going to be real.
<em>Answer; The discriminate is > 0 so there will be 2 real different roots.</em>