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yulyashka [42]
3 years ago
5

3x^2 y^7 • 5y^8 x^4 someone help plz

Mathematics
1 answer:
ale4655 [162]3 years ago
5 0

Answer:

15x^6y^15

Other:

Brainliest? Thanks!

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 Just do 7 x 5 = 35.............   
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Simplify 7/8 divided by 8
meriva

Answer:

0.109375

Step-by-step explanation:

7/8 divided by 8 will be 0.109375

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Translate the following word sentence into a numerical expression: When you square five times a number, you get three more than
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“Square five times a number” with “a number” being x is the same as saying:

5x²

“three more than the number” is the same as saying:

3 + x

Therefore the sentence “When you square five times a number, you get three more than the number.” is the same as:

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2 years ago
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Find the exact location of all the relative and absolute extrema of the function. HINT [See Examples 1 and 2.] (Order your answe
icang [17]

Answer:

  • (-1, -32) absolute minimum
  • (0, 0) relative maximum
  • (2, -32) absolute minimum
  • (+∞, +∞) absolute maximum (or "no absolute maximum")

Step-by-step explanation:

There will be extremes at the ends of the domain interval, and at turning points where the first derivative is zero.

The derivative is ...

  h'(t) = 24t^2 -48t = 24t(t -2)

This has zeros at t=0 and t=2, so that is where extremes will be located.

We can determine relative and absolute extrema by evaluating the function at the interval ends and at the turning points.

  h(-1) = 8(-1)²(-1-3) = -32

  h(0) = 8(0)(0-3) = 0

  h(2) = 8(2²)(2 -3) = -32

  h(∞) = 8(∞)³ = ∞

The absolute minimum is -32, found at t=-1 and at t=2. The absolute maximum is ∞, found at t→∞. The relative maximum is 0, found at t=0.

The extrema are ...

  • (-1, -32) absolute minimum
  • (0, 0) relative maximum
  • (2, -32) absolute minimum
  • (+∞, +∞) absolute maximum

_____

Normally, we would not list (∞, ∞) as being an absolute maximum, because it is not a specific value at a specific point. Rather, we might say there is no absolute maximum.

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3 years ago
500×64+19÷2=___________
ira [324]
PEMDAS
answer: 32009.5
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3 years ago
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