Hi! Let me help you!
Before I answer your question, I should first introduce you to triangles! There ae three triangles: Isosceles triangle, equilateral triangle, and scalene triangle. If I remember it right, isosceles triangles have at least one line of symmetry; an equilater triangle must have three lines of symmetry; a scalene triangle HAS NO line of symmetry.
Now, to answer your question: the answer is FALSE. Scalene triangles have no line of symmetry.
I hope this helped you!
Answer:
![e=\frac{1}{4} fg](https://tex.z-dn.net/?f=e%3D%5Cfrac%7B1%7D%7B4%7D%20fg)
Step-by-step explanation:
Joint variation problem solve using the equation y = kxz.
e ∝ fg
e=kfg
now substitute the values
![4=k*2*8\\4=16k\\k=\frac{4}{16} \\k=\frac{1}{4}](https://tex.z-dn.net/?f=4%3Dk%2A2%2A8%5C%5C4%3D16k%5C%5Ck%3D%5Cfrac%7B4%7D%7B16%7D%20%5C%5Ck%3D%5Cfrac%7B1%7D%7B4%7D)
The relationship will be:
![e=kfg\\e=\frac{1}{4} fg](https://tex.z-dn.net/?f=e%3Dkfg%5C%5Ce%3D%5Cfrac%7B1%7D%7B4%7D%20fg)
Answer:
Option 4 is correct
Step-by-step explanation:
If the rate is compounded continuously, the formula used to find the future value is:
A= Pe^rt
Where A = Future Value
P= Principal amount
r = interest rate in decimal
t = time
For the given data:
A=?
P = $5000
r = 7% or 0.07
t = 6
Putting values in the above formula
A= 5000e^(0.07 *6)
A = 7609.81
So, Option 4 is correct.
Answer:
distance = 9.495 mm
Step-by-step explanation:
Given data
slits distance = 0.100 mm
viewing screen distance = 1.50 m
wavelength = 633 nm
to find out
how far apart are the bright fringes on the viewing screen
solution
we use here double slit diffraction condition that is
d sin(θ) = m λ .......................1
here m = 0, 1 , 2 , 3 .....
and here for θ small so
sin(θ) = tan(θ) = ![\frac{x}{l}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7Bl%7D)
so x = l sin (θ)
so from equation 1
x =
and
x1 =
...............2
x2 =
..............3
so distance x2 - x1 will be
distance = ![\frac{\lambda l}{d}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Clambda%20l%7D%7Bd%7D)
distance = ![\frac{633*10^{-9}*1.5}{0.1*10^{-3}}](https://tex.z-dn.net/?f=%5Cfrac%7B633%2A10%5E%7B-9%7D%2A1.5%7D%7B0.1%2A10%5E%7B-3%7D%7D)
distance = 9.495 mm
Answer:
6.28 mm
Step-by-step explanation:
We have been given the diameter of circle shape necklace
![Radius=\frac{diameter}{2}](https://tex.z-dn.net/?f=Radius%3D%5Cfrac%7Bdiameter%7D%7B2%7D)
![Radius=\frac{4}{2}=2](https://tex.z-dn.net/?f=Radius%3D%5Cfrac%7B4%7D%7B2%7D%3D2)
So, the radius of circle shape necklace is 2mm
Using the formula for circumference which is:![2\pi\cdot r](https://tex.z-dn.net/?f=2%5Cpi%5Ccdot%20r)
Circumference of necklace of circle shaped is:
![2\cdot3.14\cdot2](https://tex.z-dn.net/?f=2%5Ccdot3.14%5Ccdot2)
![\Rightarrow 12.56mm](https://tex.z-dn.net/?f=%5CRightarrow%2012.56mm)
And as we are given that the circumference of of the earrings is one-half the circumference of the necklace.
Hence, the the circumference of the circle-shaped part of one of the earrings:
![\frac{1}{2}\cdot 12.56](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ccdot%2012.56)
![\Rightarrow 6.28](https://tex.z-dn.net/?f=%5CRightarrow%206.28)