Answer:
=> equation 1
=> equation 2
=> equation 3
Step-by-step explanation:
Points P, Q, R, S, and T are collinear therefore, the following equations can be written based on the segment addition postulate:
=> equation 1
=> equation 2
=> equation 3
More equations can actually be written from the diagram given using the segment addition postulate. Such as:
Your answer is C c:
First, split the angle into two angles where the values of the six trigonometric functions are known. In this case, 105, can be split into 45+60 (cos (45+60))
Use the sum formula for cosine to simplify the expression. The formula states that cos (A+B) = - (cos (A) cos (B) + sin (A) sin (B)).
The exact value of cos(60) is 1/2. So multiply 1/2 to cos (45) - sin (60) times sin (45).
The exact value of cos (45) is the square root of 2 over 2
(1/2) ⋅ (<span>√2/</span>2) - sin (60) <span>⋅ sin (45)
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Value of sin (60) is <span>√3/2
</span>(1/2) ⋅ (√2/2) - √3/2 ⋅ sin (45)
exact value of sin (45) is √2/2
(1/2) ⋅ (√2/2) - √3/2 ⋅ √2/2
all that equals √2/4 - √6/4
Answer:
um what
Step-by-step explanation:
Answer:
Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²
Step-by-step explanation:
Feets of fencing = 14
Perimeter = 14
Let x = Length y = width
Perimeter = 2(x + y)
For x = 1
2(1 + y) = 14
1 + y = 7
y =6
Area of rectangle ; x *y = 1 * 6 = 6 ft²
For x = 2
2(2 + y) = 14
2 + y = 7
y = 5
Area of rectangle ; x *y = 2 * 5 = 10 ft²
For x = 3
2(3 + y) = 14
3 + y = 7
y =4
Area of rectangle ; x *y = 3 * 4 = 12 ft²
For x = 4
2(1 + y) = 14
4 + y = 7
y = 3
Area of rectangle ; x *y = 4 * 3 =12 ft²
Hence, Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²
Use the general formula:
Volume = Height * (area of top+4*area at mid height+area at bottom) /6
Example for a cube of side s:
Vcu=s(s^2+4s^2+s^2)/6=s^3
Example for a cone of radius r
Vco=h(0+4*pi r^2/4+pi r^2)/6=pi r^2 h/3
Example for a sphere or radius r:
Vsph=2r*(0+4pi r^2 + 0)/6 = 4pi r^2 /3
So for a square frustrum with end sides a & b, height h:
Vf=h*(a^2+4((a+b)/2)^2+b^2)/6=(a^2+(a+b)^2+b^2)h/6
or simply
Vf=(a^2+(a+b)^2+b^2)h/6
