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r-ruslan [8.4K]
3 years ago
6

What is the greatest common factor of 9x^2 2 and 6x?

Mathematics
1 answer:
zheka24 [161]3 years ago
7 0

Answer:

The greatest common factor is 1.

Step-by-step explanation:

In this question we will factorize the expressions given and select the common factors among all.

For 9x² = 1×3×3×X×X

For 2 = 1×2

For 6x = 1×2×3×X

So we find the common factor in all expressions is = 1

Answer is 1.

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I'm a little bit confused about how to approach this can someone help?
Taya2010 [7]

Answer:

4

Step-by-step explanation:

ΔABC is congruent with ΔDEF.  That means the corresponding sides are congruent.

Look at the order of the letters.  EF are the last two letters of ΔDEF.  So this corresponds with the last two letters of ΔABC.  Therefore, EF ≅ BC.

EF ≅ BC

x² + 8x = 48

x² + 8x − 48 = 0

(x + 12) (x − 4) = 0

x = -12 or 4

Since x can't be negative, x = 4.

5 0
3 years ago
Chapter 4: Explain how you can
xxTIMURxx [149]

x-intercept(s):  

( 7 , 0 )   , (-1,0)

y-intercept(s):  

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Step-by-step explanation:

3 0
2 years ago
Pavel, Ruth and Josh have part-time jobs.
MrMuchimi
M+m+z/3=3m
m+m+z=9m
2m+z=9m
z=7m
m+m+7m/3=3m
9m=9m
m=1
1+1+z/3=3
2+z=9
z=7
idk but i think that one person gets paid 7 dollars and the other ppl get paid 1 dollar----- this is probably veryyyyy wrong


8 0
3 years ago
Heyy! i’ll give brainliest please help.
san4es73 [151]

Answer:

The answer is B.

Step-by-step explanation:

Its B because the lower the altitude the higher the pressure, and in this case the air towards the ground is being moved up, which is letting cold air sink back down.

8 0
3 years ago
Matlab the equation of a circle with its center at x=3 and y=2 is given by , where r is the radius of the circle. first derive t
Blizzard [7]
The equation of a circle with center at (h,k) and radius r is
(x-h)^2+(y-k)^2=r^2
we are given
center is at (3,2)
(x-3)^2+(y-2)^2=r^2
solving for y
(y-2)^2=-1(x-3)^2+r^2
y-2=\sqrt{-1(x-3)^2+r^2}
y=2+\sqrt{-1(x-3)^2+r^2}
take the derivitive to find the slope at any point
\frac{dy}{dx}=((-1)(x-3)^2+r^2))(-x+3)


we can use point slope form
for apoint (h,k) and slope m, the equation is
y-k=m(x-h)
not sure if you want in terms of what
I'll just say for a point (h,k)
so we get
y-k=((-1)(x-3)^2+r^2))(-x+3)(x-h)
where k=2+\sqrt{-1(h-3)^2+r^2}
that's the equation of the tangent line at (h,k) for arbitrary values of r

do the plots and other stuff yourself
7 0
3 years ago
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