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3 years ago

Without using a calculator, find the largest prime factor of 15^4 + 2 * 15^2 + 1 - 14^4?

Mathematics

2 answers:

fgiga [73]3 years ago

7 0

The answer is 12660, i remember i has this question before

nexus9112 [7]3 years ago

3 0

The answer is 12660

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What is the value of n in 2x3xnxn=54

Snowcat [4.5K]

54/(2X3)=54 divided by 6, =9 square root of 9 is 3

n=3

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3 years ago

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Which of the following is the equation of a line in slope-intercept form for a

Alina [70]

C is the correct answer!

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3 years ago

PLEASE HELP ME OUT LOOK AT THE IMAGE BELOW

Lelu [443]

Answer:

Let me know if you want more explanation

Step-by-step explanation:

Alright, to solve for x we need 2 angles that are congruent or equal.

I'm going to use the two we will need later. KLN and MNL

Now to find x ^^

$2x-18=x+6\\2x-18-x=x+6-x\\x-18=6\\x-18+18=6+18\\x=24$

Now we just need to plug it into each angle. I'm going to show you them separately.

$KLN=2x-18\\KLN=2(24)-18\\KLN=48-18\\KLN=30$

$MNL=4x-6\\MNL=4(24)+6\\MNL=102$

should be right now.

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3 years ago

Simplify the following

Bond [772]

Answer:

Step-by-step explanation:

1) 5xy - xy + 2x - 13x + 10xy - 8x =<u> 5xy - xy + 10xy</u> +<u>2x - 13x - 8x</u>

Combine like terms

= (5 -1 +10)xy + (2-13-8)x

= 14xy + (-19x)

= 14xy - 19x

2) -6c - (-6c) - (+6c) = -6c + 6c - 6c

= (- 6 + 6 - 6)c

= -6c

3) 0x - 22x - 4(2x) = 0x - 22x - 8x = -30x

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3 years ago

Simplify the expression ((81y^-8)/(16x^20))^-1/4

Darina [25.2K]

$\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad 
a^n\implies \cfrac{1}{a^{-n}}
\\\\
-------------------------------$

$\bf \left( \cfrac{81y^{-8}}{16x^{20}} \right)^{-\frac{1}{4}}\implies \left( \cfrac{16x^{20}}{81y^{-8}} \right)^{\frac{1}{4}}\implies \left( \cfrac{2^4x^{20}}{3^4y^{-8}} \right)^{\frac{1}{4}}
\\\\\\
\textit{and now we'll distribute the exponent}\qquad \left( \cfrac{2^{4\cdot \frac{1}{4}}x^{20\cdot \frac{1}{4}}}{3^{4\cdot \frac{1}{4}}y^{-8\cdot \frac{1}{4}}} \right)
\\\\\\
\cfrac{2^1x^5}{3^1y^{-2}}\implies \cfrac{2^1x^5y^2}{3^1}\implies \cfrac{2x^5y^2}{3}$

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3 years ago