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

2 answers:

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!

You might be interested in

(3/5y^4)^-2 simplify this please

Usimov [2.4K]

$(\frac{3}{5}y^4)^{-2}=(\frac{5}{3})^2y^{4\times-2}=\frac{25}{9y^8}$

8 0

3 years ago

8.4 gallons of gas for $23.94. Which equation can be used to determine the price of each gallon of gas?

mojhsa [17]

Here is the equation: (gas=g)

8.5g= 23.94

All you need to do is the inverse of multiplication on the other side. (division)

23.94/8.5= 2.81647.....

g= 2.82 (rounded to the hundredths place)

I hope this helps!
~kaikers

6 0

3 years ago

Please show me how to solve the initial value problem <br> y'=tanx y(pi/4)=3

AlexFokin [52]

The ODE is separable, i.e. you can write

$\dfrac{\mathrm dy}{\mathrm dx}=\tan x\iff\mathrm dy=\tan x\,\mathrm dx$

Integrating both sides gives the general solution.

$\displaystyle\int\mathrm dy=\int\tan x\,\mathrm dx$
$y=-\ln|\cos x|+C$

Given that $y\left(\dfrac\pi4\right)=3$ , we have

$3=-\ln\left|\cos\dfrac\pi4\right|+C$
$3=-\ln\dfrac1{\sqrt2}+C$
$3-\ln\sqrt2=C$

and so the particular solution to the IVP is

$y=-\ln|\cos x|+3-\ln\sqrt2$
$y=3-\ln|\sqrt2\cos x|$