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Lady bird [3.3K]
3 years ago
10

If=y=1 z=2(2y + z) (2y-z)​

Mathematics
1 answer:
snow_tiger [21]3 years ago
5 0

Answer:

answer is 0

Step-by-step explanation:

(2×1+2)(2×1-2)

(4×0)

0

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Answer:

Step-by-step explanation:

\sf Sin \ F = \dfrac{Opposite \ of \ angle \ F}{hypotenuse}

       = \dfrac{9}{18}\\\\= 0.5

    F = Sin^{-1} \ (0.5)\\

        = 30°

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2 years ago
In circle o, diameter AB, chord BC, and radius OC
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I think it's the 180 one
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Find the slope for the line through each pair of points (-18,-20) (-18,-15) NEED ASAP
nirvana33 [79]

Answer:

Distance

5

Midpoint

(−18,−17.5)

Slope

Inf∈ity

Step-by-step explanation:

no solution

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3 years ago
Read 2 more answers
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2 years ago
Factor the polynomial, x2 + 5x + 6
patriot [66]

Answer:

Choice b.

x^{2} + 5\, x + 6 = (x + 3)\, (x + 2).

Step-by-step explanation:

The highest power of the variable x in this polynomial is 2. In other words, this polynomial is quadratic.

It is thus possible to apply the quadratic formula to find the "roots" of this polynomial. (A root of a polynomial is a value of the variable that would set the polynomial to 0.)

After finding these roots, it would be possible to factorize this polynomial using the Factor Theorem.

Apply the quadratic formula to find the two roots that would set this quadratic polynomial to 0. The discriminant of this polynomial is (5^{2} - 4 \times 1 \times 6) = 1.

\begin{aligned}x_{1} &= \frac{-5 + \sqrt{1}}{2\times 1} \\ &= \frac{-5 + 1}{2} \\ &= -2\end{aligned}.

Similarly:

\begin{aligned}x_{2} &= \frac{-5 - \sqrt{1}}{2\times 1} \\ &= \frac{-5 - 1}{2} \\ &= -3\end{aligned}.

By the Factor Theorem, if x = x_{0} is a root of a polynomial, then (x - x_0) would be a factor of that polynomial. Note the minus sign between x and x_{0}.

  • The root x = -2 corresponds to the factor (x - (-2)), which simplifies to (x + 2).
  • The root x = -3 corresponds to the factor (x - (-3)), which simplifies to (x + 3).

Verify that (x + 2)\, (x + 3) indeed expands to the original polynomial:

\begin{aligned}& (x + 2)\, (x + 3) \\ =\; & x^{2} + 2\, x + 3\, x + 6 \\ =\; & x^{2} + 5\, x + 6\end{aligned}.

4 0
2 years ago
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