Answer:
the price will decrease by 4148
the price of the next model will be 29852
Answer: C) 50
Step-by-step explanation:
The smallest number on the plot is 43. The largest is 93. The range of a chart is the largest number - the smallest number. Thus, simply do 93-43 to get 50.
Hope it helps <3
Answer:
Step-by-step explanation:
se the graph to determine the input values that
correspond with f(x) = 1.
O x=4
O x= 1 and x = 4
O x= -7 and x = 4
O x= -7 and x = 2
6.
(-6, 4)
4
(1,4)
w
2
(-7, 1)
(2, 1) x
2
4
-8/ -6 -4 -2
-2
-4
Answer:
![cos(\theta)=-\frac{\sqrt{35} }{6}](https://tex.z-dn.net/?f=cos%28%5Ctheta%29%3D-%5Cfrac%7B%5Csqrt%7B35%7D%20%7D%7B6%7D)
Step-by-step explanation:
Recall the negative angle identity for the sine function:
Then, we can find the value of
:
![sin(\theta)=-sin(-\theta)\\sin(\theta) =-(-\frac{1}{6} )\\sin(\theta)= \frac{1}{6}](https://tex.z-dn.net/?f=sin%28%5Ctheta%29%3D-sin%28-%5Ctheta%29%5C%5Csin%28%5Ctheta%29%20%3D-%28-%5Cfrac%7B1%7D%7B6%7D%20%29%5C%5Csin%28%5Ctheta%29%3D%20%5Cfrac%7B1%7D%7B6%7D)
Now recall the definition of the tangent function:
![tan(\theta)=\frac{sin(\theta)}{cos(\theta)}](https://tex.z-dn.net/?f=tan%28%5Ctheta%29%3D%5Cfrac%7Bsin%28%5Ctheta%29%7D%7Bcos%28%5Ctheta%29%7D)
Therefore, now that we know the value of
, we can solve in this equation for ![cos(\theta)](https://tex.z-dn.net/?f=cos%28%5Ctheta%29)
![tan(\theta)=\frac{sin(\theta)}{cos(\theta)}\\-\frac{\sqrt{35} }{35} =\frac{1/6}{cos(\theta)} \\cos(\theta)=-\frac{\frac{1}{6} }{\frac{\sqrt{35} }{35} } \\cos(\theta)=-\frac{35}{6\,\sqrt{35} } \\cos(\theta)=-\frac{\sqrt{35} }{6}](https://tex.z-dn.net/?f=tan%28%5Ctheta%29%3D%5Cfrac%7Bsin%28%5Ctheta%29%7D%7Bcos%28%5Ctheta%29%7D%5C%5C-%5Cfrac%7B%5Csqrt%7B35%7D%20%7D%7B35%7D%20%3D%5Cfrac%7B1%2F6%7D%7Bcos%28%5Ctheta%29%7D%20%5C%5Ccos%28%5Ctheta%29%3D-%5Cfrac%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%7B%5Cfrac%7B%5Csqrt%7B35%7D%20%7D%7B35%7D%20%7D%20%5C%5Ccos%28%5Ctheta%29%3D-%5Cfrac%7B35%7D%7B6%5C%2C%5Csqrt%7B35%7D%20%7D%20%5C%5Ccos%28%5Ctheta%29%3D-%5Cfrac%7B%5Csqrt%7B35%7D%20%7D%7B6%7D)
Answer:
(1 ) Inner curved surface area of the well is 109.9 sq. meters.
(2) The cost of plastering the total curved surface area is 4396.
Step-by-step explanation:
The inner diameter = 3.5 m
Depth of the well = 10 m
Now, Diameter = 2 x Radius
⇒R = D/ 2 = 3.5/2 = 1.75
or, the inner radius of the well = 1.75 m
CURVED SURFACE AREA of cylinder = 2πr h
⇒The inner curved surface area = 2πr h = 2 ( 3.14) (1.75)(10)
= 109 sq. meters
Hence, the inner curved surface area of the well is 109.9 sq. meters.
Now, the cost of plastering the curved area is 40 per sq meters
So, the cost of total plastering total area = 109.9 x(Cost per meter sq.)
= 109.9 x (40)
= 4396
Hence, the cost of plastering the total curved surface area is 4396.