Answer:
7(2a +7)
Step-by-step explanation:
Use algebra for such problems
let, Angle COD = x
Angle KOD = y
Angle KPC = z
Given ,. x - y = 61° ( Equation 1 )
x - z = 53° ( Equation 2 )
Subtract 1st equation from 2nd and you'll get :
z - y = 8° ( Equation 3 )
Now since , x + y + z = 180° ( Equation 4 )
Add Equation 3 to Equation 4 and you'll get
x + 2z = 188° ( Equation 5)
From Equation 2 we know that , x -z = 53°
or, x = 53° + z ( Equation 6 )
Put this value of 'x' in Equation 5 and solve for z. you'll get :
(53° + z) + 2z = 188°
or
3z = 188 - 53 = 135°
solving for z we get
z = 45°
put this value of z in Equation 5
x + ( 2 x 45° ) = 188°
or
x = 188° - 90° = 98°
hence , Angle COD = 98°
Hello.
-7h + 2(-4h + 5) > -4h + 1 + 10 ; solve for h
Our first step is to get rid of all the parenthesis. We can do this by simplifying everything out.
-7h - 8h + 10 > -4h + 11
Add like terms.
-15h + 10 > -4h + 11
Now, we must isolate our variables. We do this by moving all variables to one side and moving our other numbers to the other.
Subtract 10 from both sides. Then, add 4h to both sides.
-15h + 4h > 11 - 10
Simplify.
-h > 1
Divide both sides by -1.
However, keep in mind that when dividing with a negative, you must change the sign. Therefore, > turns into <
-h ÷ -1 < 1 ÷ -1
h < -1
Answer:
90k-20k = 70k they had 1709 interest money
Step-by-step explanation:
Answer:
56.39 nm
Step-by-step explanation:
In order to have constructive interference total optical path difference should be an integral number of wavelengths (crest and crest should be interfered). Therefore the constructive interference condition for soap film can be written as,
where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...
Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.
In order to have its longest constructive reflection at the red end (700 nm)
Here we take m=0.
Similarly for the constructive reflection at the blue end (400 nm)
Hence the thickness difference should be
