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zheka24 [161]
3 years ago
10

What is the greatest number of obtuse angles a triangle can contain?

Mathematics
2 answers:
NikAS [45]3 years ago
5 0
I think 1 because if you had 2 you'll end up with a shape that is greater than 180 degrees. Keep in mind obtuse angles a over 90 degrees. hope that helped
Gennadij [26K]3 years ago
3 0
Only one. Obtuse angles are greater than 90 degrees, and a triangle has interior angles that equal 180 degrees in total, so any more than one means interior angles greater than 180. Hope this helps!
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Answer:

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Step-by-step explanation:

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Anestetic [448]

Arrange in ascending order:-

3, 5, 23, 27, 35, 37, 45, 48, 49

There are 9  values so the median = middle 5th value

=  35  Answer

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Bas_tet [7]

Answer:

y = 4

Step-by-step explanation:

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What are the limits of integration if the summation the limit as n goes to infinity of the summation from k equals 1 to n of the
Fofino [41]

Answer:

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Step-by-step explanation:

We want to express \lim_{n\rightarrow \infty} \sum_{k=1}^n\frac{7}{n}(2+\frac{7k}{n})^2 as a integral. To do this, we have to identify \sum_{k=1}^n\frac{7}{n}(2+\frac{7k}{n})^2 as a Riemann Sum that approximates the integral. (taking the limit makes the approximation equal to the value of the integral)

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In particular, if we select the ending point of each subinterval as the y_k, the Riemann sum is:

\sum_{k=1}^n f(a+k\frac{b-a}{n}) \frac{b-a}{n}.

Now, let's identify this in \sum_{k=1}^n\frac{1}{7n}(2+\frac{7k}{n})^2 .

The integrand is x² so this is our function f. When k=n, the summand should be \frac{b-a}{n}f(b)=\frac{b-a}{n}b^2 because the last selected point is b. The last summand is \frac{7}{n}(9)^2 thus b=9 and b-a=7, then 9-a=7 which implies that a=2.

To verify our answer, note that if we substitute a=2, b=9 and f(x)=x² in the general Riemann Sum, we obtain the sum inside the limit as required.

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