Lets calculate linear equation
assume
y = f(x)= mx +b
now given
f(0) = 2
m×0+b = 2
b = 2
so b = 2
given f(4) = 0.125 = 1/8
m×4 + 2 = 1/8
4m = 1/8-2= -15/8
m = -15 /(8×4) = -15/32
so linear equation is
y = mx + b = -15x/32 + 2
exponential equation is same as you have written
y = 2(0.5)^x
now put values of x in each equation to get value of y
x =0
linear y = -15×0/32+2 = 2
exp y = 2(0.5)^0 = 2×1 = 2
x =1
linear y = -15×1/32+2 = 2-15/32= 49/32
exp y = 2(0.5)^1 = 2×1/2 = 1
x =2
linear y = -15×2/32+2 = 2-30/32= 17/16
exp y = 2(0.5)^2 = 2×1/4 = 0.5
x =3
linear y = -15×3/32+2 = 2-45/32= 19/32
exp y = 2(0.5)^3 = 2×1/8= 0.25
x =4
linear y = -15×4/32+2 = 2-60/32= 4/32 = 0.125
exp y = 2(0.5)^4= 2×1/16 = 1/8= 0.125
(2,5) is not a solution because 2(2)+2(5)=15 is not true.
The answer to the question you asked is option D
Answer:
is always an odd number
Step-by-step explanation:
100-45=55
50-23=27
2-1=1
Answer:
Step-by-step explanation:
The picture of the question in the attached figure
step 1
Find the value of x
Let
O ----> the center of the circle
we know that
Triangle BOC≅Triangle COD
----> by central angle
----> by central angle
therefore
substitute the given values
solve for x
step 2
Find the measure of angle BAD
we know that
The inscribed angle is half that of the arc it comprises.
so
![m\angle BAD=\frac{1}{2} [arc\ BC+arc\ CD]](https://tex.z-dn.net/?f=m%5Cangle%20BAD%3D%5Cfrac%7B1%7D%7B2%7D%20%5Barc%5C%20BC%2Barc%5C%20CD%5D)
substitute
![m\angle BAD=\frac{1}{2} [73^o+73^o]=73^o](https://tex.z-dn.net/?f=m%5Cangle%20BAD%3D%5Cfrac%7B1%7D%7B2%7D%20%5B73%5Eo%2B73%5Eo%5D%3D73%5Eo)
