Which type of angles could be used as a counterexample for the argument all angles are either right obtuse or acute

Mathematics

1 answer:

Anettt [7]2 years ago

4 0

I believe it would be a right angle

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Can 12(x^3+y^2)+6(y^2+1) be the factored form of 12x^3+18y^2+6?

Misha Larkins [42]

Answer:

YES

Step-by-step explanation:

So how you can figure this out is you do the Distributive Property.

So it will be 12x^3+12y^2+6y^2+6.

The next thing you have to do is add together the ones that are similar which is 12y^2+6y^2 to get 18y^2.

Your answer is 12x^3+18y^2+6. So YES it can be factored into that form.

4 0

3 years ago

Read 2 more answers

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notsponge [240]

Probability=blue marbles/total marbles

Probability=5/10=1/2

If you draw from the bag twice...

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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

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