First find the decimal equivalent of square root 3: SQRT(3) = 1.732 ( roughly)
If the base and height were each 3, then the hypotenuse would need to be:
3^2 + 3^2 = C^2
9 + 9 = C^2
18 = C^2
C = SQRT(18) = 4.24
This is larger than sqrt(3), so this cannot be a right triangle.
If one leg was 3 and the other leg was sqrt(3) then the hypotenuse would be:
3^2 + 1.73^2 = C^2
9 + 3 = C^2
12 = C^2
C = SQRT(12) = 3.46
This is larger than 3, this cannot be a right triangle.
The answer is b) no.
For this, there is two rules. The common slope-intercept form is y=ax + b. For parallel, b can change, but a, or the slope, can't change. But for perpendicular, the a should be -1/a, where the answer for times the original equation, or a, and the second equation, or -1/a, is -1. To prove these two rules, you can graph it using random numbers but follow these two rules. if you have any part that don't understand for this answer, feel free to ask in the "Ask for details" section.
Answer:
Yes
Step-by-step explanation:
So I'll assume that x3 is 3x -->
so 3x • 3x • 3x =? to 3x • 3 • 3?
You can first simplify this to 3^3x =? to 3x • 9 -->
27x =? to 3x • 9 => so there are many solutions but if per say x = 5 then yes it would be equal.
Check with another number per instance 8
27(8) = 216 => 24(9) = 126
Yes they are equal
Hope this helps!
