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yaroslaw [1]
3 years ago
12

Sqrt(2x-6)=3-x Solve for x

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
5 0

Answer:

x = 3

Step-by-step explanation:

\sqrt{2x - 6}  = 3 - x

Square both sides of the equation

2x - 6 = (3 - x)^{2} = 9 - 6x + x^{2} \\

x^{2}  - 8x + 15 = 0\\

(x - 3)(x - 5) = 0

x = 3 or 5

Now, you must always check your results because a result may not satisfy the original equation.

If x = 3, then \sqrt{2x - 6} = \sqrt{2(3) - 6} = \sqrt{6 - 6}  = \sqrt{0} = 0  and 3 - x = 3 - 3 = 0

So 3 satisfies the original.

If x = 5, then \sqrt{2(5) - 6}  = \sqrt{10 - 6} = \sqrt{4} = 2,  but 3 - x = 3 - 5 = -2.  Therefore, 5 does NOT satisfy the original equation.

That means that x = 3 is the solution to the equation.

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UkoKoshka [18]

Answer:

y = -2·x^2 - 5·x - 18

y = -2·(x^2 + 10·x + 36)

y = -2·(x^2 + 10·x + 36)

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Maybe there is a typo in the original question, because there is no factorisation in the real numbers.

Because the question is " find the minimum value of y" the coefficient before x^2 must be a positive number. so it could be

y = 2·x^2 - 5·x - 18

Can you check this out?

4 0
3 years ago
At what point will 3y=4x-6 and 8x-6y=-30 intercept
tester [92]

Answer:

The lines don't intercept

Step-by-step explanation:

we have

3y=4x-6

isolate the variable y

Divide by 3 both sides

y=\frac{4}{3}x-\frac{6}{3}

simplify

y=\frac{4}{3}x-2 -----> equation A

8x-6y=-30

Isolate the variable y

6y=8x+30

Divide by 6 both sides

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simplify

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Compare equation A and equation B

The slopes are the same and the y-intercepts are different

Remember that

If two lines has the same slope, then the lines are parallel

therefore

In this problem line A and line B are parallel lines

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3 0
4 years ago
What is the equivilent of +367 on a numberline?
Mama L [17]

Answer:

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Step-by-step explanation:

367 because it is positive

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The radius of a cone is 4.9 cm and its height is 10 cm. Find the volume of the cone to the
AlladinOne [14]

Answer:

V = 251.3

Step-by-step explanation:

<u>Volume of a Cone formula:</u>

V = πr²(h/3)

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