Answer
given,
work = 5 J
spring stretch form 30 cm to 39 cm
x = 0.39 - 0.30 = 0.09 m
k = 1234.568 N/m
a) work when spring is stretched from 32 cm to 34 cm
x₂= 0.34 -0.30 = 0.04 m
x₁ = 0.32 - 0.30 = 0.02
W = 0.741 J
b) F = k x
25 = 1234.568 × x
x = 0.0205 m
x = 2.05 cm
Answer:
A dilation has center (0,0). Find the image of the point L (-5,0) for the scale factor 4.
Step-by-step explanation:
A dilation has center (0,0). Find the image of the point L (-5,0) for the scale factor 4.
So lets get to the problem
<span>165°= 135° +30° </span>
<span>To make it easier I'm going to write the same thing like this </span>
<span>165°= 90° + 45°+30° </span>
<span>Sin165° </span>
<span>= Sin ( 90° + 45°+30° ) </span>
<span>= Cos( 45°+30° )..... (∵ Sin(90 + θ)=cosθ </span>
<span>= Cos45°Cos30° - Sin45°Sin30° </span>
<span>Cos165° </span>
<span>= Cos ( 90° + 45°+30° ) </span>
<span>= -Sin( 45°+30° )..... (∵Cos(90 + θ)=-Sinθ </span>
<span>= Sin45°Cos30° + Cos45°Sin30° </span>
<span>Tan165° </span>
<span>= Tan ( 90° + 45°+30° ) </span>
<span>= -Cot( 45°+30° )..... (∵Cot(90 + θ)=-Tanθ </span>
<span>= -1/tan(45°+30°) </span>
<span>= -[1-tan45°.Tan30°]/[tan45°+Tan30°] </span>
<span>Substitute the above values with the following... These should be memorized </span>
<span>Sin 30° = 1/2 </span>
<span>Cos 30° =[Sqrt(3)]/2 </span>
<span>Tan 30° = 1/[Sqrt(3)] </span>
<span>Sin45°=Cos45°=1/[Sqrt(2)] </span>
<span>Tan 45° = 1</span>
<span>The formula for the volume of a box is:
Volume = L*W*H
We know that the length is 10 cm.
We know that the width is twice the height, so W = 2H.
Then our volume equation is:
Volume = 10*2H*H
Volume = 20*H^2
Since H^2 is in the equation, this is a quadratic equation.</span>
Answer:
2ax − 6ay + bx − 3
Step-by-step explanation:
The expression is not factorable with rational numbers
