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3 years ago

Determine the value of x if (211) base x = 152 base 8

1 answer:

8 0

It's the same as x^211 = 8^152

Simplify that equation. Find the 211th root of (8^152), and that is what x equals. If I remember logs correctly...

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The expression P(4, 3) is equal to:<br> -3!<br> -4!<br> - 1

nexus9112 [7]

Answer:

$P(4,3) = 4!$

Step-by-step explanation:

Given

P(4,3)

Required

Solve

Using permutation formula;

$P(n,r) = \frac{n!}{(n-r)!}$

This implies that

$P(4,3) = \frac{4!}{(4-3)!}$

$P(4,3) = \frac{4!}{(1)!}$

$P(4,3) = \frac{4!}{1!}$

$P(4,3) = \frac{4!}{1}$

$P(4,3) = 4!$

6 0

2 years ago

Is the question asking to see if x = 1, y=-2, and z=2 by solving the three equations?

dybincka [34]

Answer:

2 × 1 +2 +2×2 =8

2+2+4=8

8=8

3×1 +3× -2 + 4×2 =5

3-6+8=5

5=5

1+2 × -2 +2 = -5

1-4+2= -5

-5 = -5

6 0

2 years ago

A boat makes a 120-mile trip downstream in 3 hours but makes the return trip in 4 hours.

Free_Kalibri [48]

Answer: C 5mph

Hope this helps!!

6 0

3 years ago

PLEASE HELP ITS A TEST!!

Gnoma [55]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Find the simple interest and amount when principal = Rs. 6000, rate = 6% per annum and time = 4 years

Alinara [238K]

Given:

Principal = Rs. 6000

Rate of simple interest = 6% per annum.

Time = 4 years

To find:

The simple interest and amount.

Solution:

Formula for simple interest:

$I=\dfrac{P\times r\times t}{100}$

Where, P is principal, r is the rate of interest and t is the number of years.

Putting P=6000, r=6 and t=4, we get

$I=\dfrac{6000\times 6\times 4}{100}$

$I=60\times 6\times 4$

$I=1440$

Now,

$Amount=Principal+Interest$

$Amount=6000+1440$

$Amount=7440$

Therefore, the simple interest is Rs. 7440 and the amount is Rs 7440.

7 0

2 years ago