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zepelin [54]
4 years ago
14

Solve the radical equation q-6=√27-2q

Mathematics
1 answer:
rewona [7]4 years ago
5 0

q - 6 = sqrt(27 - 2q)

Squaring both sides:-

(q - 6)^2 = 27 - 2q

q^2 - 12q + 36 = 27 - 2q

q^2 - 10q + 9 = 0

(q - 9)(q - 1) = 0


so there are 2 solutions q = 1 and q = 9.


Testing to see if the solutions are not extraneous:-


q = 9:-

9 - 6 = 3

sqrt(27-18) = 3 So q = 9 is a solution (answer)


q = 1

q - 6 = -5

sqrt (27 - 2) = sqrt 25 = -5 so q = 1 is also an answer

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Select the correct answer from each drop-down menu. Quadrilateral PQRS, with vertex P(-5, -3), undergoes a transformation to for
anygoal [31]

Answer:

i) A. 180º rotation about the origin, ii) Q' = (4, 5).

Step-by-step explanation:

i) In this case, we understand that vertex P = (-5,-3) changed to P' = (5,3) after doing an operation. At first we must calculate the distance of each point regarding origin by Pythagorean Theorem:

Point P:

OP = \sqrt{(x_{P}-x_{O})^{2}+(y_{P}-y_{O})^{2}}

If we know that x_{P} = -5, y_{P} = -3, x_{O} = 0 and y_{O} = 0, the distance OP is:

OP = \sqrt{(-5-0)^{2}+(-3-0)^{2}}

OP \approx 5.831

Point P':

OP' = \sqrt{(x_{P'}-x_{O})^{2}+(y_{P'}-y_{O})^{2}}

If we know that x_{P'} = 5, y_{P'} = 3, x_{O} = 0 and y_{O} = 0, the distance OP' is:

OP' = \sqrt{(5-0)^{2}+(3-0)^{2}}

OP' \approx 5.831

As OP = OP', origin is the center of rotation.

Besides, P' is a multiple of P, that is:

1) (-5, -3) Given

2) ((-1)\cdot 5, (-1)\cdot 3) (-a)\cdot b = -a\cdot b

3) (-1)\cdot (5, 3) Scalar multiplication of a vector/Result.

The value of the scalar proves that P experimented a 180º rotation about the origin. Hence, the correct answer is A.

ii) If Q = (-4, -5) and the same operation described in item i) is used, then, the location of Q' is:

Q' = (-1)\cdot Q

Q' = (-1) \cdot (-4,-5)

Q' = ((-1)\cdot (-4), (-1)\cdot (-5))

Q' = (4, 5)

Which corresponds to option C.

7 0
3 years ago
What is answer to x+10=-13  what does x equal
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CHECK

-23 + 10 = -13
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Answer

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Step-by-step explanation:

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44.1

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