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4 years ago

Use the properties of exponents to simplify the expression all the way.

Mathematics

1 answer:

Anna71 [15]4 years ago

7 0

${( {2}^{1} {x}^{4} {y}^{ - 3} ) }^{ - 1} \: = \: ({ {2}^{1 \times ( - 1)} {x}^{4 \times ( - 1)} {y}^{ ( - 3) \times ( - 1)} ) }^{ \frac{ - 1}{ - 1} } \: = \: ( {2}^{ - 1} {x}^{ - 4} {y}^{3} ) \: = \: \frac{ {y}^{3} }{2 {x}^{4} }$

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Represent √10 , √17 and √13 in a number line.

Contact [7]

Answer:

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I I I

√10 √13 √17

Step-by-step explanation:

√10 is the smallest value, √13 is the middle value, and √17 is the largest value.

3 0

3 years ago

Three students solve a challenge math problem. Every day, the number of students who solve the problem doubles. There are 384 st

attashe74 [19]

OK, so this is assuming we are considering that the first three kids to solve ARE NOT in the first day.

So we have 3 kids and it doubles

Day 1 : 6

Day 2 : 12

Day 3 : 24

Day 4 : 48

Day 5 : 96

Day 6 : 192

Day 7: 384

So it should take 7 days or a week to solve all the problems.

The equation:

(3 * 2)^x = 384

8 0

3 years ago

Read 2 more answers

Your answer<br> What is the value of the expression? -50 + 51? *

bija089 [108]

Answer:

The answer is 1

Step-by-step explanation:

-50+50=1

3 0

3 years ago

Read 2 more answers

What is the length, in units, of the hypotenuse of a right triangle if each of the two legs is three units?

aleksandrvk [35]

It is equal to 18 and find the square root of that...........

8 0

4 years ago

Approximate the sum of the convergent infinite series the least usingnumber of terms so that the error is less than .0001. What

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Solution :

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ $a_n= \frac{(-1)^n}{n ! 2^n} \ \ \ \ \ a_{n+1}= \frac{(-1)^{n+1}}{(n+1) ! 2^{n+1}}$

Error = $|a_{n+1}|$

Error ≤ 0.0001

$|\frac{(-1)^{n+1}}{(n+1)!2^{n+1}}| \leq 0.0001$

$|\frac{1}{(n+1)!2^{n+1}}| \leq 10^{-4}$

$(n+1)! 2^{n+1} \geq 10000$

Now try, n ≥ 5

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = $\sum_{n=1 }^{5 } \frac{(-1)^n}{n ! 2^n}$ (with error 0.0001)

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = $\frac{-1}{1!2}+\frac{1}{2!2^2}-\frac{1}{3! 2^3}+\frac{1}{4! 2^4}-\frac{1}{5! 2^5}$

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = 0.6065104

3 0

3 years ago