Answer:
x = ( k+c)/5
Step-by-step explanation:
5x - c = k
Add c to each side
5x -c+c = k+c
5x = k+c
Divide by 5
5x/5= ( k+c)/5
x = ( k+c)/5
Answer:
B)
Step-by-step explanation:
- Rational number: can be expressed as a ratio of two integers
- Irrational number: cannot be expressed as a ratio of two integers



F(x)=(-2/((x+y-2)^(1/2))-(x+y+2)^(1/2)
the only irrational part of this expression is the (x+y-2)^(1/2) in the denominator, so, to rationalize this, you multiply the numerator and denominator by the denominator, as well as the other parts of the expression
also, you must multiply the -sqrt(x+y+2) by sqrt(x+y-2)/sqrt(x+y-2) to form a common denominator
(-2)/(x+y-2)^(1/2)-(x+y+2)^(1/2)(x+y-2)^(1/2)/(x+y-2)^(1/2)
(common denominator)
(-2-(x^2+xy+2x+xy+y^2+2y-2x-2y-4))/(x+y-2)^(1/2)
(FOIL)
(-2-x^2-y^2-2xy+4)/(x+y-2)^(1/2)
(Distribute negative)
(-x^2-y^2-2xy+2)/(x+y-2)^(1/2)
(Simplify numerator)
(-x^2-y^2-2xy+2)(x+y-2)^(1/2)/(x+y-2)^(1/2)(x+y-2)^(1/2)
(Rationalize denominator by multiplying both top and bottom by sqrt)
(-x^2-y^2-2xy+2)((x+y-2)^(1/2))/(x+y-2)
(The function is now rational)
=(-x^2-y^2-2xy+2)(sqrt(x+y-2))/(x+y-2)
Answer:
The general plan is to find BM and from that CM. You need 2 equations to do that.
Step One
Set up the two equations.
(7 - BM)^2 + CM^2 = (4*sqrt(2) ) ^ 2 = 32
BM^2 + CM^2 = 5^2 = 25
Step Two
Subtract the two equations.
(7 - BM)^2 + CM^2 = 32
BM^2 + CM^2 = 25
(7 - BM)^2 - BM^2 = 7 (3)
Step three
Expand the left side of the new equation labeled (3)
49 - 14BM + BM^2 - BM^2 = 7
Step 4
Simplify And Solve
49 - 14BM = 7 Subtract 49 from both sides.
-49 - 14BM = 7 - 49
- 14BM = - 42 Divide by - 14
BM = -42 / - 14
BM = 3
Step Five
Find CM
CM^2 + BM^2 = 5^2
CM^2 + 3^2 = 5^2 Subtract 3^2 from both sides.
CM^2 = 25 - 9
CM^2 = 16 Take the square root of both sides.
sqrt(CM^2) = sqrt(16)
CM = 4 < Answer
Step-by-step explanation:
Vertical angles are congruent, so their measures are equal.
Set the angle measures equal and solve for x.
2x + 2 = 3x - 52
-x = -54
x = 54
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| 54 |
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The answer is inside a box.