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cestrela7 [59]
3 years ago
15

Which choices are real numbers. Check all that apply. A. (-1024)^1/4. B. (-131072)^1/16. C. (-256)^1/9. D. (-531441)^1/13

Mathematics
2 answers:
olganol [36]3 years ago
6 0

Answer:

B and C

Step-by-step explanation:

Because it is raised to the ^Power of 13. This choice makes the most sense.

vredina [299]3 years ago
4 0

Answer:

B

Step-by-step explanation:

That is my final answer so I maybe wrong so don't take my word for it

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Nick's boss called to ask if he could cover another employee's shift on Friday night. However, Nick said that he was busy becaus
earnstyle [38]

Answer: $25

Step-by-step explanation:

8 0
3 years ago
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What is the slope-intercept form of 12x + 4y = 32?
valentina_108 [34]

Answer:

D. y=-3x+8

Step-by-step explanation:

first, get the 12x on the other side of the equation

12x+4y-12x=32-12x

4y=-12x+32

divide the equation by four

y=-3x+8

Hope this helps!

5 0
2 years ago
WILL GIVE BRAINLIEST---
Sergeeva-Olga [200]
It's a rational number because 25 is perfect square, taking the square root gives the rational integers +5 or -5
3 0
3 years ago
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Need help fast please
romanna [79]

Answer:

Step-by-step explanation:

#1 is 24k^15

just add the #

#2 4096

4x4x4=64^2

64x64

=4096

4 0
3 years ago
EXAMPLE 1 (a) Find the derivative of r(t) = (2 + t3)i + te−tj + sin(6t)k. (b) Find the unit tangent vector at the point t = 0. S
Tatiana [17]

The correct question is:

(a) Find the derivative of r(t) = (2 + t³)i + te^(−t)j + sin(6t)k.

(b) Find the unit tangent vector at the point t = 0.

Answer:

The derivative of r(t) is 3t²i + (1 - t)e^(-t)j + 6cos(6t)k

(b) The unit tangent vector is (j/2 + 3k)

Step-by-step explanation:

Given

r(t) = (2 + t³)i + te^(−t)j + sin(6t)k.

(a) To find the derivative of r(t), we differentiate r(t) with respect to t.

So, the derivative

r'(t) = 3t²i +[e^(-t) - te^(-t)]j + 6cos(6t)k

= 3t²i + (1 - t)e^(-t)j + 6cos(6t)k

(b) The unit tangent vector is obtained using the formula r'(0)/|r(0)|. r(0) is the value of r'(t) at t = 0, and |r(0)| is the modulus of r(0).

Now,

r'(0) = 3t²i + (1 - t)e^(-t)j + 6cos(6t)k; at t = 0

= 3(0)²i + (1 - 0)e^(0)j + 6cos(0)k

= j + 6k (Because cos(0) = 1)

r'(0) = j + 6k

r(0) = (2 + t³)i + te^(−t)j + sin(6t)k; at t = 0

= (2 + 0³)i + (0)e^(0)j + sin(0)k

= 2i (Because sin(0) = 0)

r(0) = 2i

Note: Suppose A = xi +yj +zk

|A| = √(x² + y² + z²).

So |r(0)| = √(2²) = 2

And finally, we can obtain the unit tangent vector

r'(0)/|r(0)| = (j + 6k)/2

= j/2 + 3k

8 0
3 years ago
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