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yuradex [85]
3 years ago
7

(25 POINTS) How do I structure this properly? I don't understand the way they are wanting me to do this, but I got the right ans

wer which is 0.61 * 10^3... Please answer with the proper answer and step..1.1 *10^3-4.9*10^2

Mathematics
2 answers:
sergij07 [2.7K]3 years ago
8 0

See the attached picture for solution and steps.

pickupchik [31]3 years ago
7 0

Answer:

1.1\times 10^3-4.9\times 10^2=6.1\cdot 10^2

Step-by-step explanation:

Given : Expression  1.1\times 10^3-4.9\times 10^2

To find : Arrange the steps in correct order to perform this subtraction problem ?

Solution :

Step 1 - Write the expression,

1.1\times 10^3-4.9\times 10^2

Step 2 - Multiply and divide by 10,

1.1\times 10^3-4.9\times \frac{10}{10}\times 10^2  

Step 3 - Solve the exponent,

1.1\times 10^3-0.49\times 10^3    

Step 4 - Taking common term,

(1.1-0.49)\cdot 10^3    

Step 5 - Subtract the bracket,

0.61\cdot 10^3  

Step 6 - Multiply and divide by 10 to convert into scientific form,

\frac{10}{10}\cdot 0.61\cdot 10^3

Step 7 - Write in proper form,

6.1\cdot 10^2

Therefore, 1.1\times 10^3-4.9\times 10^2=6.1\cdot 10^2

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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]

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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]

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