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Evgesh-ka [11]
3 years ago
15

Which of these is equal to 4 (−)4 A.()4 B.(−)4 .−()4 D. (−)4

Mathematics
1 answer:
Gennadij [26K]3 years ago
5 0

Answer:

(a) x^4 * (-y)^4 = (xy)^4

Step-by-step explanation:

Given

x^4 * (-y)^4

Required

Select and equivalent expression

A.\ (xy)^4     B.\ (-xy)^4   C.\ -(xy)^4    D.\ (x-y)^4

x^4 * (-y)^4

Apply law of indices:

x^4 * (-y)^4 =x^4 * (-y) * (-y) * (-y) * (-y)

This gives:

x^4 * (-y)^4 = x^4 * (y) * (y) * (y) * (y)

x^4 * (-y)^4 =x^4 * y^4

x and y have the same exponent; So, they can be expressed as:

x^4 * (-y)^4 = (x*y)^4

x^4 * (-y)^4 = (xy)^4

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Two children own two-way radios that have a maximum range of 3 miles. One leaves a certain point at 1:00 P.M., walking due north
Grace [21]

Answer:

The answer is 21 minutes

Step-by-step explanation:

We use the equation Xf = Xo + vt

1) At 1:00 PM, child one leaves the starting point heading north at a constant velocity of 6 mi/hr or .1 [mi/min] (divide by 60 to convert from [mi/hr] to [mi/min])

2) He walks for 15 minutes before kid 2 starts walking. In 15 minutes he is able to cover 1.5 [mi]

  • x_{1f1} =x_{o} +v_{1} t_{1} \\x_{1f1} =0+.1(15)\\x_{1f1} =1.5 [mi]

3) Now, child 2 starts walking and we know that when the range reaches 3 miles, they won´t be able to communicate. So the sum of the final position of child 1 and child 2 must be 3[mi]

  • Child 1 final position => x_{1f} = x_{1f1} +v_{1} t\\x_{1f} =1.5+v_{1} t
  • Child 2 final position => x_{2f} =0+v_{2} t

4) Sum the equations and equate to 3

  • x_{1f} +x_{2f} =3

5) Substitute the values we already know

  • 1.5+v_{1} t+v_{2}t=3\\ 1.5+.1t+.15t=3\\1.5+.25t=3\\t=\frac{3-1.5}{.25} \\t=6 [min]

6) in 15 + 6 minutes they will be 3miles apart

7) In 21 minutes they will still be able to communicate with one another.

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steposvetlana [31]

Answer:

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Step-by-step explanation:

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