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Kaylis [27]
3 years ago
15

Consider the composite function (g (x) ) = StartRoot StartFraction 1 Over x squared minus 3 EndFraction EndRoot. If g(x) = x2 –

3, what is f(x)?
f (x) = StartRoot StartFraction negative 1 Over x EndFraction EndRoot
f (x) = StartRoot StartFraction negative 1 Over x + 6 EndRoot EndFraction
f (x) = StartRoot StartFraction 1 Over x EndFraction EndRoot
f (x) = StartRoot StartFraction 1 Over x + 6 EndFraction EndRoot
Mathematics
1 answer:
balandron [24]3 years ago
8 0

Answer:

answer is c

Step-by-step explanation:

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Solve for x in the equation x^2-10x+25=35.
KatRina [158]
So, this is too vague but I'll solve for both quadratic formula and find the discriminant.

Quadratic formula: <span>x=<span>5+<span><span><span>√35 OR</span><span> </span></span>x</span></span></span>=<span>5−<span>√<span>35
Finding the discriminant: 141</span></span></span>
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MIRCrence.
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2) A gym charges a $30 membership fee and $10 per month. Write an
kari74 [83]

C = 10m + 30, where C is total cost and m is # of months

C = 10(5) + 30

C = 50 + 30

C = $80

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Hope that helps

5 0
2 years ago
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)

We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>

The vector product pq x pr gives a vector perpendicular to both pq and pr.  This vector is the normal vector of a plane passing through all three points
pq x pr
=
  i   j   k
-4 -2 -4
-3  5  1
=<-2+20,12+4,-20-6>
=<18,16,-26>

Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>

The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).

The equation of the required plane is therefore
Π :  9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π  :  9x+8y-13z=24

Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.

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3 years ago
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Step-by-step explanation:

the answer as shown in the photo

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