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Ratling [72]
3 years ago
15

What is the solution set of x + 5x -5 = 0?

Mathematics
1 answer:
Nikolay [14]3 years ago
8 0

Answer:

5/6

Step-by-step explanation:

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A line, y = mx + b, passes through the point (1, 6) and is parallel to
koban [17]

Answer: b=6

Step-by-step explanation:

6 is in the y intercept place so it equals 6

8 0
3 years ago
Solve the equation x-20.7=9.5 for x
kenny6666 [7]
The answer will come out to 30.2
8 0
2 years ago
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12x−5y=−20 y=x+4 x=x=x, equals y=y=y, equals
aksik [14]

Answer:

The solution is (0, 4)

Step-by-step explanation:

Please pay attention to the first two equations and drop the last two:

12x−5y=−20 y=x+4 x=x=x, equals y=y=y   should ideally be:

12x−5y=−20

y=x+4

Let's find x.  Substitute x + 4 for y in the first equation, obtaining:

12x - 5(x + 4) = -20

Carrying out the indicated multiplication, we get:

12x - 5x - 20 = -20, or 7x = 0

If x = 0 then y must be 0 + 4, or 4.  

The solution is (0, 4)

8 0
3 years ago
Find the difference.<br> (9/4x + 6)-(-5/4x-24)
Svetlanka [38]

Answer:

9x/4 + 30 + 5x/4

Step-by-step explanation:

9x/4 +30 +5x/5

4 0
3 years ago
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Let a = x2 + 4. Use a to find the solutions for the following equation:
Zepler [3.9K]

The solutions for x are -2, 0, 2

<em><u>Solution:</u></em>

Given that,

\text { Let } a = x^2 + 4

Given equation is:

(x^2 + 4)^2 + 32 = 12x^2 + 48

(x^2 + 4)^2 + 32 = 12(x^2 + 4)

\text { Subtsitute } a = x^2 + 4 \text{ in above equation }

a^2 + 32 = 12a\\\\a^2 -12a + 32 = 0

\text{ Let us factorize the above equation }\\\\\text{ Splitting the middle term -12a as -4a - 8a we get, }

a^2 -4a - 8a + 32 = 0

Taking "a" as common term from first two terms and taking "-8" as common from last two terms

a(a-4)-8(a - 4) = 0

\text{Taking (a - 4) as common term, }\\\\(a - 4)(a - 8) = 0

\text{Equating to zero we get, }\\\\(a - 4) = 0 \text{ or } (a - 8) = 0\\\\a = 4 \text{ or } a = 8

\text{Now substitute the value of a = 8 and a = 4 in: }\\\\a = x^2 + 4

\text{ For a = 8: }\\\\a = x^2 + 4\\\\8 = x^2 + 4\\\\x^2 = 4\\\\x = \pm2\\\\x = +2 \text{ or } -2

\text{For a = 4: }\\\\a = x^2 + 4\\\\4 = x^2 + 4\\\\x = 0

\text{Thus solutions of } x \text{ are: }\\\\x = 0 \text{ or } x = 2 \text{ or } x = -2

8 0
3 years ago
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