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Sergeeva-Olga [200]
2 years ago
10

What is the solution set of (x - 2)(x - 3) = 2? {1, 4} {2, 3} {4, 5}

Mathematics
2 answers:
inna [77]2 years ago
5 0
<span>(x - 2)(x - 3) = 2

x^2 - 2x - 3x + 6 - 2 = 0
x^2 - 5x + 4 = 0
(x -4)(x - 1) = 0
x - 4 = 0; x = 4
x - 1 = 0; x =1

{1,4}

answer
</span><span>{1, 4}</span>
Delvig [45]2 years ago
5 0

(x - 2)(x - 3) = 2? {1, 4} {2, 3} {4, 5}


(1,4)

A



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Hitung jarak di antara titik P(7, 1) dengan titik Q(7, 8) ​
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Answer:

7 units

Step-by-step explanation:

Distance between 2 points

=\sqrt{(y1 - y2)^{2}  + (x1 - x2) {}^{2} }

Thus, distance PQ

=\sqrt{(8 - 1)^{2}  + (7 - 7)^{2} }  \\ = \sqrt{7^{2} }  \\  = 7

Or you could simply take 8-1= 7 since both points have the same x-coordinate. Thus, PQ is a vertical line and the distance between them is due to the difference in their y-coordinate.

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The cost for a different taxi company is expressed with the equation y = 1.65x + 2.35, where x represents the miles driven and y
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Jared is placing chairs around of rectangular stage. The stage is 10 feet long and 22 feet wide. He wants to put a chair at each
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Solve each of the following initial value problems and plot the solutions for several values of y0. Then describe in a few words
Taya2010 [7]

Answer:

Step-by-step explanation:

Answer:

a) y-8 = (y₀-8)  , b) 2y -5 = (2y₀-5)

Explanation:

To solve these equations the method of direct integration is the easiest.

a) the given equation is

          dy / dt = and -8

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We change variables

          y-8 = u

         dy = du

We replace and integrate

           ∫ du / u = ∫ dt

           Ln (y-8) = t

We evaluate at the lower limits t = 0 for y = y₀

          ln (y-8) - ln (y₀-8) = t-0

Let's simplify the equation

           ln (y-8 / y₀-8) = t

           y-8 / y₀-8 =

            y-8 = (y₀-8)

b) the equation is

            dy / dt = 2y -5

            u = 2y -5

            du = 2 dy

            du / 2u = dt

We integrate

             ½ Ln (2y-5) = t

We evaluate at the limits

            ½ [ln (2y-5) - ln (2y₀-5)] = t

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c) the equation is very similar to the previous one

             u = 2y -10

             du = 2 dy

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We evaluate

             ln (2y-10) –ln (2y₀-10) = 2t

               2y-10 = (2y₀-10)

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gladu [14]
Your equation is like so A x 2 x 3 = 132
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A x 2 x 3 = 132
           /3     /3
A x 2 = 44
    /2     /2
A = 22.
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