Help me please I need answers

Mathematics

1 answer:

Naddika [18.5K]3 years ago

4 0

Answer:

give me a minute... if i dont find an answer i will delete my comment

Step-by-step explanation:

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Math swatch pls help

solniwko [45]

Answer:

The answer is 11.9cm

Step-by-step explanation:

The solution is in the image

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2 years ago

Evaluate each logarithm. Do not use a calculator. ln ^3 square root e^4

Alona [7]

Answer:

$\large\boxed{\ln\sqrt[3]{e^4}=\dfrac{4}{3}}$

Step-by-step explanation:

$\text{Use}\\\\\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\ln a^n=n\ln a\\\\\ln e=1\\-----------\\\\\ln\sqrt[3]{e^4}=\ln e^\frac{4}{3}=\dfrac{4}{3}\ln e=\dfrac{4}{3}\cdot1=\dfrac{4}{3}$

6 0

3 years ago

Find the 7th term of the geometric sequence whose common ratio is 1/3 and whose first term is 5.

wolverine [178]

Answer:

$t_7 =\frac{5}{729}$

Step-by-step explanation:

We are given that:

First term (a) = 5

common ratio (r) = $\frac{1}{3}$

The nth term of a geometric sequence is given as:

$t_n =ar^{n-1} \\\therefore t_7 =5\times \bigg(\frac{1}{3}\bigg) ^{7-1} \\\therefore t_7 =5\times \bigg(\frac{1}{3}\bigg) ^{6} \\\therefore t_7 =5\times \bigg(\frac{1}{3}\bigg) ^{6} \\\therefore t_7 =5\times \frac{1}{729}\\\\\huge\red{\boxed{\therefore t_7 =\frac{5}{729} }}$