PLEASE HELP AND FAST!!!

Which set of numbers can represent the side lengths, in centimeters, of a right triangle?

chubhunter [2.5K]

Answer:

The set {10 , 24 , 26} formed a right triangle

Step-by-step explanation:

* Lets explain how to check the sides lengths which formed a

right triangle

- In triangle ABC

# If AC is the longest side in length

# If (AC)² = (AB)² + (BC)²

∴ AB , BC , AC formed a right angle triangle

∴ m∠B = 90° (The angle opposite to the longest side)

∴ AC is the hypotenuse

* Now lets solve the problem

- In set 8 , 12 , 15

∵ The longest side is 15 cm

∴ (15)² = 225

∵ (8)² + (12)² = 64 + 144 = 208

∵ (15)² ≠ (8)² + (12)²

∴ The set not formed a right triangle

- In set 10 , 24 , 26

∵ The longest side is 26 cm

∴ (26)² = 676

∵ (10)² + (24)² = 100 + 576 = 676

∵ (26)² = (10)² + (24)²

∴ The set formed a right triangle

- In set 12 , 20 , 25

∵ The longest side is 25 cm

∴ (25)² = 625

∵ (12)² + (20)² = 144 + 400 = 544

∵ (25)² ≠ (12)² + (20)²

∴ The set not formed a right triangle

- In set 15 , 18 , 20

∵ The longest side is 20 cm

∴ (20)² = 400

∵ (15)² + (18)² = 225 + 324 = 549

∵ (20)² ≠ (15)² + (18)²

∴ The set not formed a right triangle

* The set {10 , 24 , 26} formed a right triangle

8 0

4 years ago

(98 pts + Brainliest) Question regarding logarithims and point slope form

bezimeni [28]

You are right in thinking that the base of the logarithm doesn't matter. It only affects the spread of the data points if you were to plot them, but would not ultimately have any effect on the slope of the line (but it would on the y-intercept).

One major discrepancy I'm noticing is in the values you found for $\ln\sin\theta$ . For example, if $\sin\theta=0.05$ , then you should have $\ln\sin\theta=\ln0.05\approx=-3.0$ . Not sure how you got -7.0, and the same goes for the rest of your table of values.

Another thing is that the provided solution suggests you take the average the first and last pairs of consecutive data points, and use these values in the slope formula to obtain the best-fit line's slope. If that's the case, then you should have

$\dfrac{\Delta\ln x}{\Delta\ln\sin\theta}=\dfrac{\frac{1.86+1.72}2-\frac{2.65+2.28}2}{\frac{-5.4-5.3}2-\frac{-7.0-6.4}2}=\dfrac{1.79-2.47}{-5.35+6.7}=\dfrac{-0.68}{1.35}\approx-0.5$

(i.e. you have to take the average of the given values, then use those averages in the $\Delta$ expressions - but this doesn't significantly affect the slope you found)

Ultimately, I think the problem is that your expression for the slope appears to be $\dfrac{\Delta\ln x}{\Delta\ln\sin\theta}$ , when the solution says it should be the reciprocal. I'm of the opinion that your slope is correct, since the experiment refers to $\theta$ (and hence $\sin\theta$ ) as the independent variable, and so $\Delta\ln\sin\theta$ would serve as the "run" and $\Delta\ln x$ would serve as the "rise".