Can someone help me with this
2 answers:
Answer:
Step-by-step explanation:
8.) For a triangles sides to make sense, you must be able to add up two values of the triangle, and the result should be more than the third side. Add the lowest values and see if the result is greater than the biggest number:
12.1 is less than the given side, 13.3, so a triangle cannot have the lengths.
10.) 6<x<22
To find the range for the third side of the triangle, you need to find how small x can be (the missing side) and you need to see how large it can be.
You need to see how small it can be because any two sides have to be greater than the third side. You also need to see how big it can be because, if it's too big, the other two sides will be less than the third side, which would make an open shape (see picture).
To find the range, first see how small. Subtract the known sides:
So, x has to be greater than 16.
x > 16
Now add the known sides:
x needs to be less than 28 for the other two sides to be greater than x:
x < 28
Insert the inequalities into a single inequality:
16 < x <28
X has to be greater than x, but less than 28.
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Take the derivative with respect to t
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
divide by w
we add sin(wt) to both sides
divide both sides by cos(wt)
OR
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be
since 2npi is just the period of cos
substituting our second soultion we get
since 2npi is the period
so the maximum value =
minimum value =
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
divide by w
we add sin(wt) to both sides
divide both sides by cos(wt)
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be
since 2npi is just the period of cos
substituting our second soultion we get
since 2npi is the period
so the maximum value =
minimum value =
Find the roots
solve
we use hmm, completing the suare
2(x²-1.5x)=4
divide both sides by 2
x²-1.5x=2
take 1/2 of linear coeiftn and square it
-1.5/2=-0.75, (-0.75)²=0.5625
add that to both sides
x²-1.5x+0.5625=2+0.5625
factor perfect squaer trinomial
(x-0.75)²=2.5625
square root both sides, remember to take positive and negative square roots
x-0.75=+/-√2.5625
add 0.75 to both sides
x=0.75+/-√2.5625
the roots are x=0.75+√2.5625 and x=0.75-√2.5625
1/a and 1/b
1/(0.75+√2.5625) and 1/(0.75-√2.5625)
if the roots of a quadratic equation are r1 and r2 then it factors to
(x-r1)(x-r2)
so then we can factor our equation to be
if we were to try and expand it, we would get
x²+0.75x-0.5
that's the simpliest equation with roots 1/a and 1/b where a and b are he roots of 2x²-3x=4
x²+0.75x-0.5 is answer
solve
we use hmm, completing the suare
2(x²-1.5x)=4
divide both sides by 2
x²-1.5x=2
take 1/2 of linear coeiftn and square it
-1.5/2=-0.75, (-0.75)²=0.5625
add that to both sides
x²-1.5x+0.5625=2+0.5625
factor perfect squaer trinomial
(x-0.75)²=2.5625
square root both sides, remember to take positive and negative square roots
x-0.75=+/-√2.5625
add 0.75 to both sides
x=0.75+/-√2.5625
the roots are x=0.75+√2.5625 and x=0.75-√2.5625
1/a and 1/b
1/(0.75+√2.5625) and 1/(0.75-√2.5625)
if the roots of a quadratic equation are r1 and r2 then it factors to
(x-r1)(x-r2)
so then we can factor our equation to be
if we were to try and expand it, we would get
x²+0.75x-0.5
that's the simpliest equation with roots 1/a and 1/b where a and b are he roots of 2x²-3x=4
x²+0.75x-0.5 is answer