Two points can lie in two different lines

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On #10 i need help to solve this. please and thank you.

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10a. $f(x) = x - 4$
   $h(x) = √(x - 5)$
   $(f - h)(6) = (6 - 4) - (√(6 - 5))$
   $(f - h)(6) = 2 - √(1)$
   $(f - h)(6) = 2 - 1$
   $(f - h)(6) = 1$

10b. $f(x) = x - 4$
   $g(x) = (1)/(x - 3)$
   $(f * g)(x) = (x - 4)((1)/(x - 3))$
   $(f * g)(x) = (x - 4)/(x - 3)$
   Domain: (-∞, 3) ∨ (3, ∞) {x|x ≠ 3}
   Interval Notation: (3 < x < 3), (3 > x > 3)

10c. $g(x) = (1)/(x - 3)$
   $h(x) = √(x - 5)$
   $(g * h)(x) = ((1)/(x - 3))(√(x - 5))$
   $(g * h)(x) = (√(x - 5))/(x - 3)$

10d. $f(x) = x - 4$
   $g(x) = (1)/(x - 3)$
   $h(x) = \sqrt[3]{x - 7}$
   $h(x) = (f * g)(x)$
   $\sqrt[3]{x - 7} = (x - 4)((1)/(x - 3))$
   $\sqrt[3]{x - 7} = (x - 4)/(x - 3)$
   $(\sqrt[3]{x - 7})^(3) = ((x - 4)/(x - 3))^(3)$
   $x - 7 = ((x - 4)^(3))/((x - 3)^(3))$
   $(x - 3)^(3)(x - 7) = (x - 4)^(3)$
   $(x^(3) - 9x^2 + 27x + 27)(x - 7) = (x^(3) - 12x^(2) + 48x - 64)$
   $x^(4) - 16x^(3) + 90x^(2) - 216x + 189 = x^(3) - 12x^(2) + 48x - 64$
   $x^(4) + 90x^(2) - 216x + 189 = 17x^(3) - 12x^(2) + 48x - 64$
   $x^(4) + 102x^(2) - 216x + 189 = 17x^(3) + 48x - 64$
   $x^(4) + 102x^(2) + 189 = 17x^(3) + 264x - 64$
   $x^(4) + 102x^(2) + 253 = 17x^(3) + 264x$
   $x^(4) - 17x^(3) + 102x^(2) - 264x + 253 = 0$
   $x = 4\ or\ 8$

Function f(x) = ax^{2}+bx+c, where a, b, and c are some constants. Define functions g and h as follows:g(x) = f(x+ 1)−f(x)
h(x) = g(x+ 1)−g(x)
Find algebraic form of h(x)
Can anyone explain how to make it step by step?

Answers

g(x) = f(x+1) - f(x)
=[ a(x+1)^2+b(x+1)+c ] - [ax^2+bx+c]
=[ a(x^2+2x+1) +bx + b + c ] - [ax^2 + bx + c]
=[ ax^2 + 2ax + a + bx + b + c ] - [ax^2 + bx + c]
= ax^2 + 2ax + a + bx + b + c - ax^2 - bx - c
= 2ax + a + b
Therefore g(x) = 2ax + a + b
h(x) = g(x+1) - g(x)
=2a (x+1) + a + b - [2ax+a+b]
=2ax + 1 + a +b - 2ax - a - b
Therefore h(x) = 1

Triangle P=27,Q=40,P=33 law of sines round measures to the nearest tenth

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For the law of sines, you would apply it in this particular problem like so:
Since P is 27, its angle is 33 and Q's length is 40; you would set it up like this
40/SinQ = 27/Sin33, multiply 40 with Sin33, then it would be 40Sin33, then divide it by 27. The result should be 40Sin33/27 = X

Which graph shows the quadratic function y = 3x^2 + 12x + 10?

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Answer with explanation:

→The quadratic function is

y = 3 x² + 12 x + 10

$y=3(x^2+4 x+(10)/(3))\n\ny=3[(x+2)^2-4+(10)/(3)]\n\ny=3[(x+2)^2-(2)/(3)]\n\ny+2=3(x+2)^2$

→Vertex of the parabola=(-2,-2),in third Quadrant.

→Drawing the Parabola,which will open upwards ,that is along positive y axis.

→ Line, x= -2 , is the line of symmetry of Quadratic function.

As,the function is Quadratic, it has either two real root or two imaginary.

But, it has two real root, which are , x=-2.816,-1.184, because,

Discriminant >0.

D=12²-4×3×10

=144 -120

D=24

The graph of the function is attached.

evaluate 5a + 3b and (5 + a)(3+B) for a=2 and b=4 how is the order of steps diffrent from the two expressions

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the first one = 10+12 = 22
second = 7x7 = 49

the order of the steps are different by the pluses and minuses
the first one had a multiply to 5 and b multiply to 3 and add them together
the second one had a plus 5 and b plus 3 and those two multiplied .

Calculate the loss on selling 50 shares of stock originally bought at 133/4 and sold at 12

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The answer is $87.50. You buy 50 shares of stock for 13 3/4 each: 50 * (13 3/4) = 50 * (13 + 3/4) = 50 * (52/4 + 3/4) = 50 * 55/4 = $687.50. You sell 50 shares of stock for 12 each: 50 * 12 = $600.00. So, the loss on selling 50 shares of stock originally bought at 133/4 and sold at 12 is: $687.50 - $600.00 = $87.50.

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