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Marta_Voda [28]

2 years ago

6

2x^16-32x^4 what is the completely factored form

1 answer:

marysya [2.9K]2 years ago

4 0

Answer:

B

Step-by-step explanation:

$~~~2x^{16} - 32x^4\\\\ = 2x^4(x^{12} -16)\\\\=2x^4\left[(x^6)^2 - 4^2 \right]\\\\=2x^4(x^6 -4)(x^6 +4)~~~~~~~~~~~~~~~~~~;[a^2 -b^2 = (a+b)(a-b)]\\\\=2x^4\left[(x^3)^2 - 2^2\right] (x^6 +4)\\\\=2x^4(x^3 -2)(x^3 +2)(x^6 +4)$

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Find the area of the figure below

alex41 [277]

Answer: 29

Step-by-step explanatio?

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3 years ago

If JKLM is a rhombus, MK = 30, NL = 13, and mZMKL = 41°, find each measure.

oksian1 [2.3K]

Answer:

$NK = 15$

$JL = 26$

$KL = 19.85$

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$\angle JML =41$

$\angle MLK = 90$

$\angle MNL =90$

$\angle KJL =41$

Step-by-step explanation:

Given

$MK = 30$

$NL = 13$

$\angle MKL = 41$

Solving (a): NK

MK is a diagonal and NK is half of the diagonal. So:

$NK = \frac{1}{2} * MK$

$NK = \frac{1}{2} * 30$

$NK = 15$

Solving (b): JL

JL is a diagonal, and it is twice of NL.

$JL = 2 * NL$

$JL = 2 * 13$

$JL = 26$

Solving (c): KL

To solve for KL, we consider triangle KNL where:

$\angle KNL = 90$

and

$KL^2 = NL^2 + NK^2$

$KL^2 = 13^2 + 15^2$

$KL^2 = 394$

$KL = \sqrt{394$

$KL = 19.85$

Solving (d - h):

To do this, we consider triangle JKN

$\angle KNL = \angle LNM = \angle MNJ = \angle JNK = 90$ -- diagonals bisect one another at right angle

Alternate interior angles are equal. So:

$\angle MKL = \angle KMJ = \angle KJL = \angle JLM = 41$

Similarly:

$\angle MKJ = \angle KML = \angle MJL = \angle JLK = 90 - 41$

$\angle MKJ = \angle KML = \angle MJL = \angle JLK = 49$

So:

$\angle JKM =49$

$\angle JML =41$

$\angle MLK = \angle MLJ + \angle JLK$

$\angle MLK = 49 + 41$

$\angle MLK = 90$

$\angle MNL =90$

$\angle KJL =41$

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