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zvonat [6]
3 years ago
8

Which expressions are equivalent to 25x4 −64? Select three options. 25x4 + 40x − 40x − 64 25x4 + 13x − 13x − 64 (5x2 + 8)(5x2 −

8) (x2 + 13)(x2 – 13) (5x2 – 8)2
Mathematics
2 answers:
Lana71 [14]3 years ago
7 0

Answer:

a, b, c

25x4 + 40x − 40x − 64, 25x4 + 13x − 13x − 64, (5x2 + 8)(5x2 − 8)

Step-by-step explanation:

exis [7]3 years ago
5 0

Answer:

The equivalent expression of 25 - 64 are "a), b) and c)"

Step-by-step explanation

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Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6
allsm [11]

Answer:

L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]

Step-by-step explanation:

Given that:

f(t)  = 12 cos (t- \dfrac{\pi}{6})

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

f(t) = 12 [cos\  t \  cos \dfrac{\pi}{6}+ sin \ t  \ sin \dfrac{\pi}{6}]

f(t) = 12 [cos \  t \ \dfrac{3}{2}+ sin  \ t  \ sin \dfrac{1}{2}]

f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)

L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ]

L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ]

L(f(t)) = 6 \sqrt{3}  \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}

L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}

L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}

L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]

7 0
3 years ago
PLease explain this to me
aliina [53]

Answer:

ok so brainly is a place where you ask questions :P

<em>hope this helps<3</em>

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2 years ago
What is144 dividend by 10?
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3 years ago
Read 2 more answers
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