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PolarNik [594]
3 years ago
6

Find X when x equals negative 15

Mathematics
1 answer:
Ann [662]3 years ago
7 0
Judging by the question you have provided I came to the conclusion that you have already solved your own problem!

If the goal is to find X when X=-15 then your answer for X should be -15!

If this is not the entire equation please post the entire one! 

Hope this helped!
-Blake
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G(t) = -2t + 3f(t) = 2t - 5<br> Find (g f)(t)
Tanzania [10]

Answer:

( g − f ) ( a ) = g ( a ) − f ( a )

( g −f ) ( a ) = g ( a ) − f ( a )

= − 3 a − 3 − a 2 − 5

= − a 2 − 3a − 8

Step-by-step explanation:

3 0
4 years ago
How many 2/15 meter pieces of wood can you cut from a piece that is 2/3 meter long?
Dmitry_Shevchenko [17]

You are cutting wood into equally sized pieces so if you know the length of the wood and the size of the cuts...

That would be...

2/3 ÷ 2/15 = 2/3 x 15/2 = 5 pieces

3 0
4 years ago
Read 2 more answers
Find the greatest common factor of 6y^3 and 10n^2
Tasya [4]
SOLUTION:

GCF = 2

= 6y^3 + 10n^2

= 2 ( 3y^3 + 5n^2 )

Hope this helps! :)
Have a lovely day! <3
7 0
3 years ago
I just want someone to proof-read these so that I have a good understanding of what I am doing (leave feedback)
Semmy [17]

Yea it's good but make sure to check your answers to see if you did anything wrong.

5 0
3 years ago
Find an equation for the plane that contains the line v = (−1, 1, 2) + t(7, 6, 2) and is perpendicular to the plane 8x + 5y − 7z
Nadusha1986 [10]
The required plane &Pi; contains the line 
L: (-1,1,2)+t(7,6,2)
means that &Pi; is perpendicular to the direction vector of the line L, namely
vl=<7,6,2>

It is also required that &Pi; be perpendicular to the plane
&Pi; 1 : 5y-7z+8=0
means that &Pi; is also perpendicular to the normal vector of the given plane, vp=<0,5,-7>.

Thus the normal vector of the required plane, &Pi; can be obtained by the cross product of vl and vp, or vl x vp:
 i  j  k
7 6  2
0 5 -7
=<-42-10, 0+49, 35-0>
=<-52, 49, 35> 
which is the normal vector of &Pi;

Since &Pi; has to contain the line, it must pass through the point (-1,1,2), so the equation of the plane is
&Pi; :  -52(x-(-1))+49(y-1)+35(z-2)=0
=>
&Pi; :  -52x+49y+35z = 171

Check that normal vector of plane is orthogonal to line direction vector
<-52,49,35>.<7,6,2>
=-364+294+70
=0   ok
6 0
4 years ago
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