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alekssr [168]
3 years ago
6

Applying Properties of Exponents In Exercise,use the properties of exponents to simplify the expression.

Mathematics
1 answer:
murzikaleks [220]3 years ago
3 0

Answer:

a)e^{3}e^{4}=e^{7}

b) (e^{3})^{4}=e^{12}

c) (e^{3})^{-2}=e^{-6}

d) e^{0}=1

Step-by-step explanation:

a) Here we can use the product of powers rule: x^{a}x^{b}=x^{a+b}

e^{3}e^{4}=e^{7}

b) In this ca we use power of a power rule: (x^{a})^{b}=x^{a*b}.

(e^{3})^{4}=e^{12}

c) We use the same rule above.

(e^{3})^{-2}=e^{-6}

d) Finally, we can use zero exponent rule: x^{0}=1.

e^{0}=1

I hope it helps!

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The ordered pair (a,b)satisfies the inequality y<br><br>need a little help​
Ymorist [56]

Answer:

If you add 5 to a, it will be greater than b

Step-by-step explanation:

This is because if you have the inequality y<x+5 a=x b=y so by adding 5 to a meaning x 5+a will still be greater than b(y) with out changing the sign or inequality.

3 0
2 years ago
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ra1l [238]

Answer:

Pretty sure it is A

Step-by-step explanation:

3 0
3 years ago
Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interv
sleet_krkn [62]

Answer:

(0, 16]

Step-by-step explanation:

∑ₙ₌₁°° (-1)ⁿ⁺¹ (x−8)ⁿ / (n 8ⁿ)

According to the ratio test, if we define L such that:

L = lim(n→∞) |aₙ₊₁ / aₙ|

then the series will converge if L < 1.

aₙ = (-1)ⁿ⁺¹ (x−8)ⁿ / (n 8ⁿ)

aₙ₊₁ = (-1)ⁿ⁺² (x−8)ⁿ⁺¹ / ((n+1) 8ⁿ⁺¹)

Plugging into the ratio test:

L = lim(n→∞) | (-1)ⁿ⁺² (x−8)ⁿ⁺¹ / ((n+1) 8ⁿ⁺¹) × n 8ⁿ / ((-1)ⁿ⁺¹ (x−8)ⁿ) |

L = lim(n→∞) | -n (x−8) / (8 (n+1)) |

L = (|x−8| / 8) lim(n→∞) | n / (n+1) |

L = |x−8| / 8

For the series to converge:

L < 1

|x−8| / 8 < 1

|x−8| < 8

-8 < x−8 < 8

0 < x < 16

Now we check the endpoints.  If x = 0:

∑ₙ₌₁°° (-1)ⁿ⁺¹ (0−8)ⁿ / (n 8ⁿ)

∑ₙ₌₁°° -(-1)ⁿ (-8)ⁿ / (n 8ⁿ)

∑ₙ₌₁°° -(8)ⁿ / (n 8ⁿ)

∑ₙ₌₁°° -1 / n

This is a harmonic series, and diverges.

If x = 16:

∑ₙ₌₁°° (-1)ⁿ⁺¹ (16−8)ⁿ / (n 8ⁿ)

∑ₙ₌₁°° (-1)ⁿ⁺¹ (8)ⁿ / (n 8ⁿ)

∑ₙ₌₁°° (-1)ⁿ⁺¹ / n

This is an alternating series, and converges.

Therefore, the interval of convergence is:

0 < x ≤ 16

Or, in interval notation, (0, 16].

4 0
3 years ago
Graph the given relation or equation and find the domain and range. Then
vredina [299]
X=1 Hopefully that helped
6 0
3 years ago
Which fraction is larger 7/16 or 51/100?
zaharov [31]
51/100 would be larger
From the first look, I already know half of 100 is fifty and 51 is greater than that
So when I looked at 7/16, I was looking for something greater than half, but 7 is greater than half of 16.
51/100 is your answer
5 0
3 years ago
Read 2 more answers
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