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4 years ago

True or false ???????????????please help

Mathematics

1 answer:

Dovator [93]4 years ago

7 0

1 True because 19^3 = 6859
2. False
3. True because 8000 ^1/3 can be written as 100^1/3 * 8^1/3
4 False
5 True because they are both = 15

If you have a scientific calculator you can check these out yourself

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3 decagrams = centgrams

notsponge [240]

Answer:

3,000

Step-by-step explanation:

Just multiply every decagram by 1,000

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3 years ago

Which is the sum of 3/10 and 1/3

Ilya [14]

Step-by-step explanation:

$\: \: \: \: \: \frac{3}{10} + \frac{1}{3} \\ \\ = \frac{3 \times 3}{10 \times 3} + \frac{1 \times 10}{3 \times 10} \\ \\ = \frac{9}{30} + \frac{10}{30} \\ \\ = \frac{19}{30} \\ \\ so \: as \: you \: see \: here \: we \: should \: put \: same \: denominator \\ \: for \: both \: fractions. \: hope \: this \: helps.. \\ good \: luck$

8 0

3 years ago

WILL GIVE BRAINLIEST!!!!!

Phoenix [80]

It’s either A or B but I would choose B as the correct answer

3 0

3 years ago

Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your

Darina [25.2K]

Answer:

Given definite integral as a limit of Riemann sums is:

$\lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

Step-by-step explanation:

Given definite integral is:

$\int\limits^7_4 {\frac{x}{2}+x^{3}} \, dx \\f(x)=\frac{x}{2}+x^{3}---(1)\\\Delta x=\frac{b-a}{n}\\\\\Delta x=\frac{7-4}{n}=\frac{3}{n}\\\\x_{i}=a+\Delta xi\\a= Lower Limit=4\\\implies x_{i}=4+\frac{3}{n}i---(2)\\\\then\\f(x_{i})=\frac{x_{i}}{2}+x_{i}^{3}$

Substituting (2) in above

$f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

Riemann sum is:

$= \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

4 0

4 years ago

I need help to find the rule.

Sonbull [250]

Lololololol. I long since forgot how to do this. rip

3 0

3 years ago

Read 2 more answers