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BlackZzzverrR [31]
3 years ago
9

Which polynomial is prime? x2 + 9 x2 – 25 3x2 – 27 2x2 – 8

Mathematics
1 answer:
Charra [1.4K]3 years ago
3 0
<h3>Answer: Choice A.  x^2+9</h3>

This is a sum of squares, which cannot be factored over the real numbers. You'll need to involve complex numbers to be able to factor, though its likely your teacher hasn't covered that topic yet (though I could be mistaken and your teacher has mentioned it).

Choice B can be factored through the difference of squares rule. Therefore, choice B is not prime.

Choice C and D can be factored by pulling out the GCF and then use the difference of squares rule afterward. So we can rule out C and D as well.

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13.9 is the right answer :)

Step-by-step explanation:

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Tell which congruence postulate or theorem you would use to show that thetriangles are congruent.
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From the given figure, we know that:

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Use the diagram to find lengths. BP is the perpendicular bisector of AC. QC is the
faust18 [17]

Answer:

The length of the side PC is 34 cm.

Step-by-step explanation:

We are given that BP is the perpendicular bisector of AC. QC is the perpendicular bisector of BD. AB = BC = CD.

Suppose BP = 16 cm and AD = 90 cm.

As, it is given that AD = 90 cm and the three sides AB = BC = CD.

From the figure it is clear that AD = AB + BC + CD

So, AB = \frac{90}{3} = 30 cm

BC = \frac{90}{3} = 30 cm

CD = \frac{90}{3} = 30 cm

Since the triangle, BPC is a right-angled triangle as \anglePBC = 90°, so we can use Pythagoras theorem in this triangle to find the length of the side PC.

Now, the Pythagoras theorem states that;

\text{Hypotenuse}^{2} = \text{Perpendicular}^{2} +\text{Base}^{2}

\text{PC}^{2} = \text{BP}^{2} +\text{BC}^{2}

\text{PC}^{2} = \text{16}^{2} +\text{30}^{2}

\text{PC}^{2} = 256+900 = 1156

\text{PC}=\sqrt{1156}

PC = 34 cm

Hence, the length of the side PC is 34 cm.

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x/520 = 5/13

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I took a test with this question on it.

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