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garik1379 [7]
3 years ago
5

How would you describe the difference between the graphs of f(x) = x^2 +4 and g(y) - y^2 +4?

Mathematics
1 answer:
s2008m [1.1K]3 years ago
3 0

Answer:

Step-by-step explanation:

The function f(x)=x^2+4 is a positive upwards opening parabola with the vertex at (0, 4), whereas

the function g(y)=y^2+4 is a positive rightwards opening parabola (sideways parabola) with the vertex at (4, 0). This means that answer to this is that g(y) is reflected over the x axis whereas f(x) is reflected over the y axis.

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Is 4286 divisible by 6
grigory [225]
No.
4286 / 6= 714.33333333333
3 0
3 years ago
What is the value of [(2/3)^0]^-3
o-na [289]

Answer:

1

Step-by-step explanation:

Easy because whenever the exponent is 0 the answer must always be 1 then the exponent on the outside it would still be 1.

5 0
3 years ago
How to find all real solutions
AlladinOne [14]
\bf \sqrt{\sqrt{x-5}+x}=5\leftarrow \textit{squaring both sides}
\\\\
\sqrt{x-5}+x=25\implies \sqrt{x-5}=25-x\leftarrow \textit{squaring both sides}
\\\\
x-5=(25-x)^2\implies x-5=625-50x+x^2
\\\\
0=x^2-51x+630\implies 0=(x-30)(x-21)

and surely you'd know what the roots are

7 0
3 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.

3 0
3 years ago
Simplify expression<br>8x-6+x-1​
charle [14.2K]

Answer:

9x -7

Step-by-step explanation:

8x-6+x-1​

Combine like terms

8x +x   -6-1

9x -7

8 0
3 years ago
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