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Serhud [2]
3 years ago
7

6-2x=2(15-x) A. No solutions B. -5 C. 5 D. All real numbers

Mathematics
1 answer:
andrezito [222]3 years ago
5 0
6-2x=2(15-x)
distribute
6-2x=30-2x
add 2x to both sides
6=30
fasle

no solution
A is answer
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Solve for x 1.x=1 and 2over 3​
kogti [31]

Answer:

<h3>x = 1 and 2/3 or x = 1 2/3</h3><h3 />

Step-by-step explanation:

Simplify both sides of equation

1 * x = 1 2/3

1 * x is 1x which is the same as x because there is one x

then for the improper fraction which is 1 2/3

to simplify improper fractions we times the whole number by the denominator

so 1 * 3 then add the numerator 2 and the denominator stays the same so we get x = 5/3

now we have x = 5/3 this is not correct because the numerator is bigger than the denominator so we just subtract the denominator from the numerator and how many times 3 goes into 5 (in this case 1) becomes the new whole number.

therefore we get x = 1 2/3

6 0
3 years ago
Simplify this equation
aleksklad [387]
A^2 - 4a + 4
hope this helps!
6 0
3 years ago
Read 2 more answers
8. Polynomial Division<br> Divide. (6x2 – 13x + 2) ÷ (3x – 2)
KengaRu [80]

Answer:

Quotient = -4

Remainder = -x+6

Step-by-step explanation:

(12+2-13x) / (3x-2)

= (-13x+14) / (3x-2)

Quotient = -4

Remainder = -x+6

4 0
4 years ago
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

         dy / y-8 = dt

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

            Ln (2y-5 / 2y₀-5) = 2t

            2y -5 = (2y₀-5)

c) the equation is very similar to the previous one

             u = 2y -10

             du = 2 dy

             ∫ du / 2u = dt

             ln (2y-10) = 2t

We evaluate

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

               2y-10 = (2y₀-10)

4 0
3 years ago
0.059 seconds into milliseconds
elena55 [62]
The answer would be 59 miliseconds
6 0
4 years ago
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