Answer:
The ratio of juice boxes remaining to juice boxes drunk = 20 / 60 = 1/3
Step-by-step explanation:
If the degree of numerator and denominator are equal, then limit will be leading coefficient of numerator divided by the
leading coefficient of denominator.
So then the limit would be 3/1 =
3.
Alternatively,
![\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}=\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}\cdot\dfrac{1/x^2}{1/x^2}=\lim_{x\to\infty}\dfrac{3+\frac6{x^2}}{1-\frac4{x^2}} = \dfrac{3+0}{1-0}=\boxed{3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cdfrac%7B3x%5E2%2B6%7D%7Bx%5E2-4%7D%3D%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cdfrac%7B3x%5E2%2B6%7D%7Bx%5E2-4%7D%5Ccdot%5Cdfrac%7B1%2Fx%5E2%7D%7B1%2Fx%5E2%7D%3D%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cdfrac%7B3%2B%5Cfrac6%7Bx%5E2%7D%7D%7B1-%5Cfrac4%7Bx%5E2%7D%7D%20%3D%20%5Cdfrac%7B3%2B0%7D%7B1-0%7D%3D%5Cboxed%7B3%7D)
Hope this helps.
X/12 = -8
Solve for x by multiplying both sides by 12
X = -96
Answer:
- C.) obtuse triangle
- 39°
- 167°
- D.) 111°
Step-by-step explanation:
1. The sum of angles in a triangle is 180°. The sum of the given angles is less than 90°, so the remaining angle must be more than 90°. When the largest angle is more than 90°, it is an obtuse angle. A triangle with an obtuse angle is classified as obtuse.
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2. The sum of the interior angles is 180°, so we have ...
x° +39° +102° = 180°
x° = 39° . . . . . . . . . . . subtract 141°
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3. The third interior angle is the supplement to the sum of the given interior angles. Likewise, x° is the supplement to the third interior angle, so its value is the same as the sum of the two given interior angles. (The rule is often stated as "an exterior angle is equal to the sum of the remote interior angles.")
x° = 130° +37°
x° = 167°
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4. Same deal as problem 3:
x° = 75° +36°
x° = 111°
We'll have to repeatedly square both sides of the equation, in order to get rid of the square roots. Squaring a first time yields
![19+\sqrt{30+\sqrt{32+x}}=25](https://tex.z-dn.net/?f=19%2B%5Csqrt%7B30%2B%5Csqrt%7B32%2Bx%7D%7D%3D25)
Move the 19 to the right hand side:
![\sqrt{30+\sqrt{32+x}}=6](https://tex.z-dn.net/?f=%5Csqrt%7B30%2B%5Csqrt%7B32%2Bx%7D%7D%3D6)
And square again:
![30+\sqrt{32+x}=36 \iff \sqrt{32+x}=6](https://tex.z-dn.net/?f=30%2B%5Csqrt%7B32%2Bx%7D%3D36%20%5Ciff%20%5Csqrt%7B32%2Bx%7D%3D6)
Square one last time:
![32+x=36 \iff x=36-32=4](https://tex.z-dn.net/?f=32%2Bx%3D36%20%5Ciff%20x%3D36-32%3D4)
Let's check the solutions: all these squaring might have created external solutions:
![\sqrt{19+\sqrt{30+\sqrt{32+4}}}=\sqrt{19+\sqrt{30+6}}=\sqrt{19+6}=\sqrt{25}=5](https://tex.z-dn.net/?f=%5Csqrt%7B19%2B%5Csqrt%7B30%2B%5Csqrt%7B32%2B4%7D%7D%7D%3D%5Csqrt%7B19%2B%5Csqrt%7B30%2B6%7D%7D%3D%5Csqrt%7B19%2B6%7D%3D%5Csqrt%7B25%7D%3D5)
So,
is a feasible solution.