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Rainbow [258]
3 years ago
5

Jalil and Victoria are each asked to solve the equation ax – c = bx + d for x. Jalil says it is not possible to isolate x becaus

e each x has a different unknown coefficient. Victoria believes there is a solution, and shows Jalil her work:
Mathematics
2 answers:
Iteru [2.4K]3 years ago
8 0
<span>solve the equation ax – c = bx + d for x:

1) Group the x terms together on the left:     ax - bx - c = d

2) Group the constant terms together:          ax - bx = c + d

3) factor out x:                                             x(a - b) = c + d

4) Divide both sides of the equation by (a - b) to obtain a formula for x:
                                                       c+d
</span>           x(a - b) = c + d   =>    x = ---------
                                                       a-b

This shows that the given equation CAN be solved for x, but there is a restriction:  a must NOT equal b, because if a-b = 0, we'd have division by zero (which is not defined).

Where is Victoria's solution?  Please share it if you want to discuss this problem further.  Thank you.

Yuri [45]3 years ago
5 0

Answer:

A , Rewrite the expression on the left using the distributive property.

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Given the function h(x)=-4+6,<br> Find the value of h(-2)
valina [46]

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-----------

If you mean h(x) = -4x+6

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2 years ago
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2 years ago
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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
3 years ago
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