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Yuliya22 [10]
3 years ago
10

Rewrite the expression.

Mathematics
1 answer:
Anvisha [2.4K]3 years ago
6 0
Log₄8 + 3 · log₄x

so the easiest way to do this is to note that these logs are separated by an addition symbol--it isn't "log₄8 + 3" times "log₄x"

log₄8
plus
3 · log₄x

for the second log, you can condense it with log properties/rules: the coefficient out front, when you condense it, becomes the exponent for the argument of your log:
3 · log₄x = log₄(x³)

so, having condensed that, your equation reads:

log₄8 + log₄(x³)

you could technically evaluate the first log, but the question wants both of these to become a single logarithm, which means you want to combine them. log properties state that if logs are being added, you can multiply their arguments (for example: logₓab = logₓa + logₓb)

you just want to apply that property to this, so you'll be multiplying your arguments 8 and x³:

log₄(8x³) is the answer, expressed as one logarithm.
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Find the solutions to y = (x – 1)2 for x = –3, 0, and 3.
Harman [31]

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if the 2 is the power of (x-1) here is ur answer

y = (x - 1 ){}^{2} \\if \: x \: is \: \: equal \: to \:  - 3  \\ y =(  - 3 - 1) {}^{2} \\  y = ( - 4) {}^{2} \\ y = 16

if \: x = 0 \\ y = (0 - 1) {}^{2}  \\y = (  - 1) { }^{2} \\ y = 1

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(-3,16) (0,1) (3,4)

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I don't know how to do this please help explain how to do it aswell as give the answer so i understand
Zinaida [17]

<u>We are given:</u>

The function: y = -16t² + 64

where y is the height from ground, t seconds after falling

<u>Part A:</u>

when the droplet would hit the ground, it's height from the ground will be 0

replacing that in the given function:

0 = -16t² + 64

16t² = 64                          [adding 16t² on both sides]

t² = 4                                [dividing both sides by 16]

t = 2 seconds                  [taking square root of both sides]

<u>Part B:</u>

for second droplet,

height from ground = 16 feet

time taken = t seconds

acceleration due to gravity = 10 m/s²

initial velocity = 0 m/s

h = ut + (1/2)at²                 [second equation of motion]

16 = (0)(t) + (1/2)(10)(t²)

16 = 5t²

t² = 16/5

t = 1.8 seconds (approx)

Therefore, the second droplet takes the least amount time to hit the ground

4 0
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