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Harman [31]
2 years ago
6

Solve the following system of equations.express your answer as an ordered pair in the format (a,b) with no spaces between the nu

mbers or symbols .
3x+4y=17
-4x-3y=-18
Mathematics
1 answer:
frosja888 [35]2 years ago
4 0

Answer: (3,2)

Step-by-step explanation:

For this set of system of equations, let's use elimination method to find x and y. Let's eliminate y first. In order to do so, the y values have to be the same. Therefore, you multiply the top equation by 3 and the bottom equation by 4.

9x+12y=51

-16x-12y=-72

Now, we can add the 2 equations together for the y to cancel out.

-7x=-21

x=3

Since we know x, we can plug it into any of the equations to find y.

3(3)+4y=17

9+4y=17

4y=8

y=2

With out x and y value, we can put them into an ordered pair. (3,2)

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j         < 48

j should be less than 48 to make the inequality true.

example: j = 47
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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]

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